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12.5: Tangential and Normal Components of Acceleration - Mathematics


Learning Objectives

  • Describe the velocity and acceleration vectors of a particle moving in space.
  • Explain the tangential and normal components of acceleration.
  • State Kepler’s laws of planetary motion.

We have now seen how to describe curves in the plane and in space, and how to determine their properties, such as arc length and curvature. All of this leads to the main goal of this chapter, which is the description of motion along plane curves and space curves. We now have all the tools we need; in this section, we put these ideas together and look at how to use them.

Motion Vectors in the Plane and in Space

Our starting point is using vector-valued functions to represent the position of an object as a function of time. All of the following material can be applied either to curves in the plane or to space curves. For example, when we look at the orbit of the planets, the curves defining these orbits all lie in a plane because they are elliptical. However, a particle traveling along a helix moves on a curve in three dimensions.

Definition: Speed, Velocity, and Acceleration

Let (vecs r(t)) be a twice-differentiable vector-valued function of the parameter (t) that represents the position of an object as a function of time.

The velocity vector (vecs v(t)) of the object is given by

[ ext{Velocity},=vecs v(t)=vecs r′(t). label{Eq1}]

The acceleration vector (vecs a(t)) is defined to be

[ ext{Acceleration},=vecs a(t)=vecs v′(t)=vecs r″(t). label{Eq2}]

The speed is defined to be

[mathrm{Speed},=v(t)=‖vecs v(t)‖=‖vecs r′(t)‖=dfrac{ds}{dt}. label{Eq3}]

Since (vecs{r}(t)) can be in either two or three dimensions, these vector-valued functions can have either two or three components. In two dimensions, we define (vecs{r}(t)=x(t) hat{mathbf i}+y(t) hat{mathbf j}) and in three dimensions (vecs r(t)=x(t) hat{mathbf i}+y(t) hat{mathbf j}+z(t) hat{mathbf k}). Then the velocity, acceleration, and speed can be written as shown in the following table.

Table (PageIndex{1}): Formulas for Position, Velocity, Acceleration, and Speed
QuantityTwo DimensionsThree Dimensions
Position(vecs{r}(t)=x(t) hat{mathbf i}+y(t) hat{mathbf j})(vecs{r}(t)=x(t) hat{mathbf i}+y(t) hat{mathbf j}+z(t) hat{mathbf k})
Velocity(vecs{v}(t)=x′(t) hat{mathbf i}+y′(t) hat{mathbf j})(vecs{v}(t)=x′(t) hat{mathbf i}+y′(t) hat{mathbf j}+z′(t) hat{mathbf k})
Acceleration(vecs{a}(t)=x″(t) hat{mathbf i}+y″(t) hat{mathbf j})(vecs{a}(t)=x″(t) hat{mathbf i}+y″(t) hat{mathbf j}+z″(t) hat{mathbf k})
Speed(|vecs{v}(t)|= sqrt{(x′(t))^2+(y′(t))^2})(|vecs{v}(t)|=sqrt{(x′(t))^2+(y′(t))^2+(z′(t))^2})

Example (PageIndex{1}): Studying Motion Along a Parabola

A particle moves in a parabolic path defined by the vector-valued function (vecs{r}(t)=t^2 hat{mathbf i}+ sqrt{5−t^2} hat{mathbf j}), where (t) measures time in seconds.

  1. Find the velocity, acceleration, and speed as functions of time.
  2. Sketch the curve along with the velocity vector at time (t=1).

Solution

  1. We use Equations ef{Eq1}, ef{Eq2}, and ef{Eq3}:

    [ egin{align*} vecs{v}(t) &= vecs{r}′(t)=2that{mathbf i}−dfrac{t}{sqrt{5-t^2}}hat{mathbf j} [4pt] vecs{a}(t) &=vecs{v}′(t)=2hat{mathbf i}−5(5−t^2)^{-frac{3}{2}}hat{mathbf j} [4pt] ||vecs{v}(t)|| &=||vecs{r}′(t)|| [4pt] &=sqrt{(2t)^2+left(-dfrac{t}{sqrt{5-t^2}} ight)^2} [4pt] &=sqrt{4t^2+dfrac{t^2}{5-t^2}} [4pt] &=sqrt{dfrac{21t^2-4t^4}{5-t^2}}. end{align*}]

  2. The graph of (vecs{r}(t)=t^2 hat{mathbf i}+ sqrt{5−t^2} hat{mathbf j}) is a portion of a parabola (Figure (PageIndex{1})).

    When (t=1), (vecs r(1) = (1)^2 mathbf{hat i} + sqrt{5-(1)^2} mathbf{hat j} quad = quad mathbf{hat i} + sqrt{4} mathbf{hat j} quad = quad mathbf{hat i} + 2 mathbf{hat j}).

    Thus the particle would be located at the point ((1, 2)) when (t =1).

    The velocity vector at (t=1) is

    [ egin{align*} vecs{v}(1) &=vecs{r}′(1)=2(1)hat{mathbf i}−frac{1}{sqrt{5-1^2}} hat{mathbf j}quad [4pt] &= quad 2hat{mathbf i}−frac{1}{2}hat{mathbf j} end{align*}]

    and the acceleration vector at (t=1) is

    [vecs{a}(1)=vecs{v}′(1)=2hat{mathbf i}−5(5 - 1^2)^{-3/2}hat{mathbf j}quad = quad 2hat{mathbf i}−frac{5}{8}hat{mathbf j}.]

    Notice that the velocity vector is tangent to the path, as is always the case.

Exercise (PageIndex{1})

A particle moves in a path defined by the vector-valued function (vecs r(t)=(t^2−3t),hat{mathbf i}+(2t−4),hat{mathbf j}+(t+2),hat{mathbf k}), where (t) measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.

Hint

Use Equations ef{Eq1}, ef{Eq2}, and ef{Eq3}.

Answer

[egin{align*}vecs v(t) &=vecs{r}'(t) =(2t-3),hat{mathbf i}+2,hat{mathbf j}+,hat{mathbf k}[4pt] vecs a(t) &=vecs v′(t) =2,hat{mathbf i} end{align*}]

[ ||vecs{r}′(t)||=sqrt{(2t-3)^2+2^2+1^2} =sqrt{4t^2-12t+14}]

The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.

To gain a better understanding of the velocity and acceleration vectors, imagine you are driving along a curvy road. If you do not turn the steering wheel, you would continue in a straight line and run off the road. The speed at which you are traveling when you run off the road, coupled with the direction, gives a vector representing your velocity, as illustrated in Figure (PageIndex{2}).

However, the fact that you must turn the steering wheel to stay on the road indicates that your velocity is always changing (even if your speed is not) because your direction is constantly changing to keep you on the road. As you turn to the right, your acceleration vector also points to the right. As you turn to the left, your acceleration vector points to the left. This indicates that your velocity and acceleration vectors are constantly changing, regardless of whether your actual speed varies (Figure (PageIndex{3})).

Components of the Acceleration Vector

We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion in the plane and in space. Recall that the unit tangent vector (vecs T) and the unit normal vector (vecs N) form an osculating plane at any point (P) on the curve defined by a vector-valued function (vecs{r}(t)). The following theorem shows that the acceleration vector (vecs{a}(t)) lies in the osculating plane and can be written as a linear combination of the unit tangent and the unit normal vectors.

Theorem (PageIndex{1}): The Plane of the Acceleration Vector

The acceleration vector (vecs{a}(t)) of an object moving along a curve traced out by a twice-differentiable function (vecs{r}(t)) lies in the plane formed by the unit tangent vector (vecs T(t)) and the principal unit normal vector (vecs N(t)) to (C). Furthermore,

[vecs{a}(t) = v'(t)vecs{T}(t) + [v(t)]^2 kappa vecs{N}(t) ]

Here, (v(t) = |vecs v(t)|) is the speed of the object and (kappa) is the curvature of (C) traced out by (vecs{r}(t)).

Proof

Because (vecs{v}(t)=vecs{r}′(t)) and (vecs{T}(t)=dfrac{vecs{r}′(t)}{||vecs{r}′(t)||}), we have (vecs v(t)=||vecs{r}′(t)||vecs{T}(t)=v(t)vecs{T}(t)).

Now we differentiate this equation:

[vecs{a}(t)=vecs{v}′(t)=dfrac{d}{dt}left(v(t)vecs{T}(t) ight)=v′(t)vecs{T}(t)+v(t)vecs{T}′(t) onumber]

Since (vecs{N}(t)=dfrac{vecs{T}′(t)}{||vecs{T}′(t)||}), we know (vecs{T}′(t)=||vecs{T}′(t)||vecs{N}(t)), so

[vecs{a}(t)=v′(t)vecs{T}(t)+v(t)||vecs{T}′(t)||vecs{N}(t). onumber]

A formula for curvature is (kappa=dfrac{||vecs{T}'(t)||}{||vecs{r}'(t)||}), so (vecs{T}'(t) = kappa ||vecs{r}'(t) || = kappa v(t) ).

This gives (vecs{a}(t)=v′(t)vecs{T}(t)+kappa (v(t))^2 vecs{N}(t).)

(square)

The coefficients of (vecs{T}(t)) and (vecs{N}(t)) are referred to as the tangential component of acceleration and the normal component of acceleration, respectively. We write (a_vecs{T}) to denote the tangential component and (a_vecs{N}) to denote the normal component.

Theorem (PageIndex{2}): Tangential and Normal Components of Acceleration

Let (vecs{r}(t)) be a vector-valued function that denotes the position of an object as a function of time. Then (vecs{a}(t)=vecs{r}′′(t)) is the acceleration vector. The tangential and normal components of acceleration (a_vecs{T}) and (a_vecs{N}) are given by the formulas

[a_{vecs{T}}=vecs a cdotvecs{T}=dfrac{vecs{v}cdotvecs{a}}{||vecs{v}||} label{Eq1B}]

and

[a_vecs{N}=vecs acdot vecs N=dfrac{||vecs v imes vecs a||}{||vecs v||}=sqrt{||vecs a||^2−{left(a_{vecs{T}} ight)^2}}. label{Eq2B}]

These components are related by the formula

[vecs{a}(t)=a_vecs{T} vecs{T}(t)+a_vecs{N}vecs{N}(t). label{Eq3B}]

Here (vecs{T}(t)) is the unit tangent vector to the curve defined by (vecs{r}(t)), and (vecs{N}(t)) is the unit normal vector to the curve defined by (vecs{r}(t)).

The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radial component of acceleration. To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed. Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration. The same holds true for non-circular paths. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point (B) in Figure (PageIndex{4}) the acceleration vector is pointing backward. This is because the car is decelerating as it goes into the curve.

The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. The tangential and normal components of acceleration are the projections of the acceleration vector onto (vecs T) and (vecs N), respectively.

Example (PageIndex{2}): Finding Components of Acceleration

A particle moves in a path defined by the vector-valued function (vecs{r}(t)=t^2,hat{mathbf i}+(2t−3),hat{mathbf j}+(3t^2−3t),hat{mathbf k}), where (t) measures time in seconds and distance is measured in feet.

  1. Find (a_vecs{T}) and (a_vecs{N}) as functions of (t).
  2. Find (a_vecs{T}) and (a_vecs{N}) at time (t=2).

Solution

  1. Let’s start deriving the velocityand acceleration functions:

    [egin{align*} vecs{v}(t) &= vecs{r}'(t) [4pt] &= 2t,hat{mathbf i}+2,hat{mathbf j}+(6t-3),hat{mathbf k} [4pt] vecs{a}(t) &= vecs{v}'(t) [4pt] &=2,hat{mathbf i}+6,hat{mathbf k} end{align*}]Now we apply Equation ef{Eq1B}: [egin{align*}a_{vecs{T}} &=dfrac{vecs{v}cdot vecs{a}}{||vecs{v} ||} [4pt] &= dfrac{ (2t ,hat{mathbf i} +2,hat{mathbf j} +(6t-3),hat{mathbf k})cdot(2,hat{mathbf i}+6,hat{mathbf k}) }{|| 2t,hat{mathbf i} + 2,hat{mathbf j} + (6t-3),hat{mathbf k} ||} [4pt] &= dfrac{4t + 6(6t-3)}{sqrt{(2t)^2 +2^2 + (6t-3)^2}} [4pt] &= dfrac{40t-18}{40t^2 - 36t+13} end{align*}] Now we can apply Equation ef{Eq2B}:

    [egin{align*} a_vecs{N} &=sqrt{||vecs{a}||^2-a_{vecs{T}}} [4pt] &= sqrt{||2,hat{mathbf i}+6,hat{mathbf k}||^2 - left( dfrac{ 40t-18 }{sqrt{40t^2-36+13}} ight)^2} [4pt] &= sqrt{ 4+36-dfrac{(40t-18)^2}{40t^2-36t+13} } [4pt] &= sqrt{dfrac{ 40(40t^2-36t+13)-(1600t^2-1440t+324) }{40t^2-36t+13} } [4pt] &= sqrt{ dfrac{196}{ 40t^2-36t+13} } [4pt] &= dfrac{14}{sqrt{40t^2-36t+13}} end{align*}]

  2. We must evaluate each of the answers from part a at (t=2):

    [egin{align*} a_{vecs{T}}(2) &= dfrac{ 40(2)-18 }{sqrt{40(2)^2 - 36(2)+13 }} [4pt] &= dfrac{80-18 }{sqrt{160-72+13}} [4pt] &= dfrac{62}{sqrt{101}} [4pt] a_{vecs{N}}(2) &= dfrac{14}{sqrt{40(2)^2 -36(2)+13 }} [4pt] &= dfrac{14}{sqrt{160-72+13}} = dfrac{140}{sqrt{101}}. end{align*}]

    The units of acceleration are feet per second squared, as are the units of the normal and tangential components of acceleration.

Exercise (PageIndex{2})

An object moves in a path defined by the vector-valued function (vecs r(t)=4t,hat{mathbf i}+t^2,hat{mathbf j}), where (t) measures time in seconds.

  1. Find (a_vecs{T}) and (a_vecs{N}) as functions of (t).
  2. Find (a_vecs{T}) and (a_vecs{N}) at time (t=−3).
Hint

Use Equations ef{Eq1B} and ef{Eq2B}

Answer

a. [egin{align*} a_vecs{T} =dfrac{vecs v(t) cdot vecs a(t)}{||vecs v(t)||}= dfrac{vecs r'(t) cdot vecs r''(t) }{||vecs r'(t)||} = dfrac{ (4,hat{mathbf i} + 2t ,hat{mathbf j}) cdot (2,hat{mathbf j}) }{||4,hat{mathbf i} + 2t ,hat{mathbf j} ||} = dfrac{4t}{sqrt{4^2 + (2t)^2}} = dfrac{2t}{sqrt{2+t^2}} end{align*}]
[egin{align*} a_vecs{N} = sqrt{||vecs a||^2-a_vecs{T}^2} =sqrt{||2,hat{mathbf j} ||^2 - left(dfrac{2t}{sqrt{2+t^2}} ight)^2} =sqrt{ 4 - dfrac{4t^2 }{2+t^2} } end{align*}]

b. [egin{align*} a_vecs{T}(−3) = dfrac{2(-3)}{sqrt{2+(-3)^2}} = dfrac{-6}{sqrt{11}}end{align*}]
[egin{align*}a_vecs{N}(−3) = sqrt{ 4 - dfrac{4(-3)^2 }{2+(-3)^2} } = sqrt{4- dfrac{36}{11}} = sqrt{dfrac{8}{11} } =dfrac{2sqrt{2}}{sqrt{11}} end{align*}]

Projectile Motion

Now let’s look at an application of vector functions. In particular, let’s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as projectile motion. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs.

First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive (y)-axis to be up, the negative (y)-axis to be down, and the positive (x)-axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton’s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from gravity, or (vecs F_g=mvecs a), where (vecs F_g) represents the force from gravity and (vecs a = -g,hat{mathbf j}) represents the acceleration resulting from gravity at Earth’s surface. The value of (g) in the English system of measurement is approximately 32 ft/sec2 and it is approximately 9.8 m/sec2 in the metric system. This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form (vecs F_g=−mg,hat{mathbf j}), as shown in Figure (PageIndex{5}).

Newton’s second law also tells us that (F=mvecs{a}), where (vecs a) represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that

[egin{align*} vecs F =vecs F_g mvecs{a} = -mg ,hat{mathbf j} vecs{a} = -g,hat{mathbf j}. end{align*}]

Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form

[vecs v'(t) = -g,hat{mathbf j} ]

By taking the antiderivative of each side of this equation we obtain

[ vecs v(t) = int -g ,hat{mathbf j}; dt = -gt,hat{mathbf j} + vecs C_1 ]

for some constant vector (vecs C_1). To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time (t=0). We call this velocity the initial velocity: (vecs v(0)=vecs v_0). Therefore, (vecs v(0)=−g(0),hat{mathbf j}+vecs C_1=vecs v_0) and (vecs C_1= vecs v_0). This gives the velocity vector as (vecs v(t)=−gt,hat{mathbf j}+vecs v_0).

Next we use the fact that velocity (vecs{v}(t)) is the derivative of position (vecs{s}(t)). This gives the equation

[vecs s'(t)=-gt,hat{mathbf j}+vecs{v}_0. ]

Taking the antiderivative of both sides of this equation leads to

[egin{align*} vecs s(t) &= int -gt,hat{mathbf j} + vecs{v}_0 ;dt [4pt] &= -dfrac{1}{2}gt^2 ,hat{mathbf j} + vecs{v}_0 t + vecs{C}_2 end{align*}]

with another unknown constant vector (vecs{C}_2). To determine the value of (vecs{C}_2), we can use the position of the object at a given time, say at time (t=0). We call this position the initial position: (vecs{s}(0)=vecs{s}_0). Therefore, (vecs{s}(0)=−(1/2)g(0)^2,hat{mathbf j}+vecs{v}_0(0)+vecs{C}_2=vecs{s}_0). This gives the position of the object at any time as

[ vecs{s}(t)=−12gt^2 ,hat{mathbf j}+vecs{v}_0 t+vecs{s}_0. ]

Let’s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle ( heta) to the horizontal, with initial speed (vecs{v}_0). How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, (vecs{s}_0=vecs{0}), as shown in Figure (PageIndex{6}).

We can rewrite the initial velocity vector in the form (vecs{v}_0= v_0 cos heta ,hat{mathbf i} + v_0 sin heta ,hat{mathbf j}). Then the equation for the position function (vecs{s}(t)) becomes

[egin{align*} vecs{s}(t) &=-dfrac{1}{2} gt^2,hat{mathbf j} + v_0 t cos heta ,hat{mathbf i} + v_0 t sin heta ,hat{mathbf j} [4pt] &= v_0 t cos heta,hat{mathbf i} + v_0 t sin heta ,hat{mathbf j} - dfrac{1}{2} gt^2,hat{mathbf j} [4pt] &= v_0 t cos heta ,hat{mathbf i} + left(v_0 t sin heta - dfrac{1}{2} gt^2 ight),hat{mathbf j}. end{align*}]

The coefficient of (hat{mathbf i}) represents the horizontal component of (vecs{s}(t)) and is the horizontal distance of the object from the origin at time (t). The maximum value of the horizontal distance (measured at the same initial and final altitude) is called the range (R). The coefficient of (hat{mathbf j}) represents the vertical component of (vecs{s}(t)) and is the altitude of the object at time (t). The maximum value of the vertical distance is the height (H).

Example (PageIndex{3}): Motion of a Cannonball

During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of 30° above horizontal and the initial speed of the cannonball is 600 ft/sec. The cliff is 100 ft above the water (Figure (PageIndex{7})).

  1. Find the maximum height of the cannonball.
  2. How long will it take for the cannonball to splash into the sea?
  3. How far out to sea will the cannonball hit the water?

Solution

We use the equation

[vecs{s}(t) = v_0 t cos heta ,hat{mathbf i} + left(v_0 t sin heta - dfrac{1}{2}gt^2 ight),hat{mathbf j} onumber]

with ( heta=30^circ ), (g=32 dfrac{ ext{ft}}{ ext{sec}^2}), and (v_0=600 dfrac{ ext{ft}}{ ext{sec}^2}). Then the position equation becomes

[egin{align*} vecs{s}(t) &= 600 t ( cos 30^circ),hat{mathbf i} + left(600t sin30^circ - dfrac{1}{2}(32)t^2 ight),hat{mathbf j} [4pt] &= 300tsqrt{3} ,hat{mathbf i} + left( 300t - 16t^2 ight),hat{mathbf j} end{align*}]

  1. The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is

    [egin{align*} vecs{v}(t) &=vecs s'(t)[4pt] &= 300 sqrt{3} ,hat{mathbf i} + (300-32t),hat{mathbf j} end{align*} ]

    Therefore, the vertical component of velocity is given by the expression (300−32t). Setting this expression equal to zero and solving for t gives (t=9.375) sec. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at (t=9.375).

    [egin{align*} vecs{s}(9.375) &=300(9.375)sqrt{3},hat{mathbf i}+(300(9.375)−16(9.375)^2),hat{mathbf j} [4pt] &=4871.39 ,hat{mathbf i}+1406.25,hat{mathbf j} end{align*}]

    Therefore, the maximum height of the cannonball is 1406.39 ft above the cannon, or 1506.39 ft above sea level.
  2. When the cannonball lands in the water, it is 100 ft below the cannon. Therefore, the vertical component of the position vector is equal to −100. Setting the vertical component of (vecs s(t)) equal to −100 and solving, we obtain

    [egin{align*} 300t-16t^2 &= -100 16t^2-300t-100 =0 4t^2-75-25 =0 [4pt] t &= dfrac{75pm sqrt{(-75)^2}-4(4)(-25) }{2(4)} [4pt] &= dfrac{75 pm sqrt{6025}}{8} [4pt] &= dfrac{75 pm 5sqrt{241}}{8} end{align*}]

    The positive value of (t) that solves this equation is approximately 19.08. Therefore, the cannonball hits the water after approximately 19.08 sec.
  3. To find the distance out to sea, we simply substitute the answer from part (b) into (vecs{s}(t)):

    [egin{align*} vecs s(19.08) &=300(19.08)sqrt{3} ,hat{mathbf i}+left(300(19.08)−16(19.08)^2 ight),hat{mathbf j}[4pt] &=9914.26,hat{mathbf i}−100.7424,hat{mathbf j} end{align*}]

    Therefore, the ball hits the water about 9914.26 ft away from the base of the cliff. Notice that the vertical component of the position vector is very close to −100, which tells us that the ball just hit the water. Note that 9914.26 feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is 100 ft above the water (the same as the altitude of the cannon).

Exercise (PageIndex{3})

An archer fires an arrow at an angle of 40° above the horizontal with an initial speed of 98 m/sec. The height of the archer is 171.5 cm. Find the horizontal distance the arrow travels before it hits the ground.

Hint

The equation for the position vector needs to account for the height of the archer in meters.

Answer

967.15 m

One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is

[vecs{s}=v_0 t cos heta ,hat{mathbf i} + left(v_0tsin heta - dfrac{1}{2}gt^2 ight),hat{mathbf j}.]

Setting the second component equal to zero and solving for (t) yields

[egin{align*} v_0 t sin heta - dfrac{1}{2}gt^2 =0 tleft(v_0 sin heta - dfrac{1}{2}gt ight) =0 end{align*}]

Therefore, either (t=0) or (t=dfrac{2v_0sin heta}{g}). We are interested in the second value of (t), so we substitute this into (vecs{s}(t)), which gives

[egin{align*} vecs{s}left(dfrac{2v_0sin heta}{g} ight) = v_0 left(dfrac{2v_0sin heta}{g} ight) cos heta ,hat{mathbf i} + left( v_0left(dfrac{2v_0sin heta}{g} ight)sin heta - dfrac{1}{2}gleft(dfrac{2v_0sin heta}{g} ight)^2 ight),hat{mathbf j} = left(dfrac{2v_0^2sin hetacos heta}{g} ight),hat{mathbf i} = dfrac{v_0^2 sin2 heta}{g},hat{mathbf i}. end{align*}]

Thus, the expression for the range of a projectile fired at an angle ( heta) is

[R=dfrac{v_0^2 sin2 heta}{g},hat{mathbf i} . ]

The only variable in this expression is ( heta). To maximize the distance traveled, take the derivative of the coefficient of i with respect to ( heta) and set it equal to zero:

[egin{align*} dfrac{d}{d heta} left( dfrac{v_0^2 sin2 heta}{g} ight) =0 dfrac{2v_0^2cos2 heta}{g} =0 heta=45^circ end{align*}]

This value of ( heta)) is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a 45° angle. The distance it travels is given by

[vecs{s}left(dfrac{2v_0 sin 45^circ}{g} ight)= dfrac{v_0^2 sin 90^circ}{g} ,hat{mathbf i} = dfrac{v_0^2}{g},hat{mathbf i} ]

Therefore, the range for an angle of 45° is (frac{v_0^2}{g}) units.

Kepler’s Laws

During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Brahe to formulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.

Theorem (PageIndex{2}): Kepler's Laws of Planetary Motion

  1. The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus of the ellipse (the law of ellipses).
  2. A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals (the law of equal areas) (Figure (PageIndex{8})).
  3. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths of their semimajor orbital axes (the Law of Harmonies).

Kepler’s third law is especially useful when using appropriate units. In particular, 1 astronomical unit is defined to be the average distance from Earth to the Sun, and is now recognized to be 149,597,870,700 m or, approximately 93,000,000 mi. We therefore write 1 A.U. = 93,000,000 mi. Since the time it takes for Earth to orbit the Sun is 1 year, we use Earth years for units of time. Then, substituting 1 year for the period of Earth and 1 A.U. for the average distance to the Sun, Kepler’s third law can be written as

[ T_p^2=D_p^3 onumber]

for any planet in the solar system, where (T_P) is the period of that planet measured in Earth years and (D_P) is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.

Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems, such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun.

Proof

Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function (vecs{r}(t)) represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as (vecs{F}=mvecs{a}), where (vecs{F}) represents the net force acting on the planet. His law of universal gravitation can be written in the form (vecs{F}=−dfrac{GmM}{||vecs{r}||^2}cdot dfrac{vecs{r}}{||vecs{r} ||}), which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude (dfrac{GmM}{||vecs{r}||^2} ) (Figure (PageIndex{9})).

Setting these two forces equal to each other, and using the fact that (vecs a(t)=vecs v′(t)), we obtain

[ mvecs v′(t)=−frac{GmM}{‖vecs r‖^2}⋅frac{vecs r}{‖vecs r‖}, onumber ]

which can be rewritten as

[ dfrac{dvecs v}{dt}=−dfrac{GM}{||vecs r||^3}vecs{r}. onumber ]

This equation shows that the vectors (dvecs{v}/dt) and (vecs r) are parallel to each other, so (dvecs {v}/dt imes vecs {r}=vecs 0). Next, let’s differentiate (vecs{r} imes vecs{v}) with respect to time:

[dfrac{d}{dt}(vecs{r} imes vecs{v})=dfrac{dvecs{r}}{dt} imes vecs v+vecs{r} imes dfrac{dvecs{v}}{dt}=vecs{v} imes vecs{v}+vecs{0}=vecs{0}. label{Eq10}]

This proves that (vecs{r} imesvecs{v}) is a constant vector, which we call (vecs C). Since (vecs r) and (vecs v) are both perpendicular to (vecs C) for all values of (t), they must lie in a plane perpendicular to (vecs C). Therefore, the motion of the planet lies in a plane.

Next we calculate the expression (dvecs{v}/dt imes vecs C):

[dfrac{dvecs{v}}{dt} imes vecs{C}=−dfrac{GM}{||vecs{r}||^3}vecs{r} imes (vecs{r} imesvecs{v})=−dfrac{GM}{||vecs r||^3}[(vecs{r} cdot vecs{v})vecs{r} - (vecs{r} cdot vecs{r})vecs{v}]. label{Eq11}]

The last equality in Equation ef{Eq10} is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for (vecs{r}cdot vecs{v}). To calculate this, we differentiate (vecs{r}cdot vecs{r}) with respect to time:

[ dfrac{d}{dt}(vecs{r}cdot vecs{r})=dfrac{dvecs{r}}{dt}cdot vecs{r}+vecs{r}cdot dfrac{dvecs{r}}{dt}=2vecs{r}cdot dfrac{dvecs{r}}{dt}=2vecs{r}cdot vecs{v}. label{Eq12}]

Since (vecs{r}cdotvecs{r}=||vecs r||^2), we also have

[dfrac{d}{dt}(vecs{r}cdot vecs{r})=dfrac{d}{dt}||vecs{r}||^2=2||vecs{r}|| dfrac{d}{dt}||vecs{r}||. label{Eq13}]

Combining Equation ef{Eq12} and Equation ef{Eq13}, we get

[egin{align*} 2vecs{r}cdot vecs{v} =2||vecs{r}||dfrac{d}{dt}||vecs{r}|| vecs{r} cdot vecs{v} =||vecs{r}‖dfrac{d}{dt}||vecs{r}||. end{align*} label{Eq14}]

Substituting this into Equation ef{Eq11} gives us

[egin{align} dfrac{dvecs{v}}{dt} imes vecs{C} = - dfrac{GM}{||vecs{r}||^3} [(vecs{r}cdot vecs{v})vecs{r} - (vecs{r}cdot vecs{r})vecs{v}] onumber = -dfrac{GM}{||vecs{r}||^3}left[ ||vecs{r} left(dfrac{d}{dt} ||vecs{r}|| ight)vecs{r} - ||vecs{r}||^2vecs{v} ight] onumber = -GMleft[ dfrac{1}{||vecs{r}||^2}left( dfrac{d}{dt} ||vecs{r}|| ight)vecs{r} - dfrac{1}{||vecs{r}||}vecs{v} ight] onumber = GMleft[ dfrac{vecs{v}}{||vecs{r}||} -dfrac{vecs{r}}{||vecs{r}||^2}left( dfrac{d}{dt} ||vecs{r}|| ight) ight]. label{Eq15} end{align}]

However,

[ egin{align*} dfrac{d}{dt} dfrac{vecs{r}}{||vecs{r}||} = dfrac{ frac{d}{dt}(vecs{r})||vecs{r}||- vecs{r}frac{d}{dt}||vecs{r}|| }{||vecs{r}||^2} = dfrac{ frac{dvecs{r}}{dt} }{||vecs{r}||} - dfrac{vecs{r}}{||vecs{r}||^2}dfrac{d}{dt}||vecs{r} || = dfrac{vecs{v}}{||vecs{r}||} - dfrac{vecs{r}}{||vecs{r}||^2} dfrac{d}{dt}||vecs{r}||. end{align*}]

Therefore, Equation ef{Eq15} becomes

[dfrac{d vecs{v}}{dt} imes vecs{C}=GMleft( dfrac{d}{dt}dfrac{ vecs{r}}{ || vecs{r} ||} ight). onumber ]

Since (vecs{C}) is a constant vector, we can integrate both sides and obtain

[ vecs{v} imesvecs{C} = GM dfrac{ vecs{r} }{|| vecs{r} ||} + vecs{D}, onumber ]

where (vecs D) is a constant vector. Our goal is to solve for (|| vecs{r} ||). Let’s start by calculating ( vecs{r} cdot ( vecs{v} imes vecs{C}):

[vecs{r} cdot ( vecs{v} imes vecs{C} =GMdfrac{||vecs{r}||^2}{||vecs{r}||}+ vecs{r}cdotvecs{D} =GM||vecs{r}||+vecs{r}cdot vecs{D}. onumber ]

However, ( vecs{r} cdot ( vecs{v} imes vecs{C})= ( vecs{r} imes vecs{v})cdot vecs{C} ), so

[ ( vecs{r} imes vecs{v})cdot vecs{C} =GM||vecs{r}|| + vecs{r}cdot vecs{D}. onumber ]

Since (vecs{r} imes vecs{v}=vecs{C}), we have

[ ||vecs{C}||^2 =GM||vecs{r}|| +vecs{r}cdot vecs{D}. onumber ]

Note that ( vecs{r} cdot vecs{D}=||vecs{r}|| ||vecs{D}||cos heta ), where ( heta) is the angle between (vecs{r}) and (vecs{D}). Therefore,

[ ||vecs{C}||^2=GM||vecs{r}||+||vecs{r}|| ||vecs{D}|| cos heta onumber ]

Solving for (||vecs{r}||),

[ ||vecs{r}|| = dfrac{||vecs{C}||^2 }{GM+||vecs{D}||cos heta} = dfrac{||vecs{C}||^2}{GM}left( dfrac{1}{1+ecos heta} ight). onumber ]

where (e=||vecs{D}||/GM). This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if (e>1), a parabola if (e=1), or an ellipse if (e<1). Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.

(square)

Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes

[P^2 = dfrac{4pi^2 a^3}{G(m+M)}, label{Eq30}]

where m is the mass of the Moon and M is the mass of Earth, a represents the length of the major axis of the elliptical orbit, and P represents the period.

Example (PageIndex{4}): Using Kepler’s Third Law for Nonheliocentric Orbits

Given that the mass of the Moon is (7.35 imes 10^{22}) kg, the mass of Earth is (5.97 imes 10^{24}) kg, (G=6.67 imes 10^{−11} ext{m} / ext{kg} cdot ext{sec}^2), and the period of the moon is 27.3 days, let’s find the length of the major axis of the orbit of the Moon around Earth.

Solution

It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:

[27.3 ; ext{days} imes dfrac{24 ; ext{hr}}{1 ; ext{day}} imes dfrac{3600 ; ext{esc}}{1 ; ext{hour}} =2,358,720; ext{sec} onumber ]

Substitute all the data into Equation ef{Eq30} and solve for (a):

[egin{align*} (2,358,720sec)^2 = dfrac{4pi^2a^3}{left( 6.67 imes 10^{-11} frac{m}{ ext{kg} imes ext{sec}^2} ight) (7.35 imes 10^{22} ext{kg} + 5.97 imes 10^{24} ext{kg})} 5.563 imes 10^{12} = dfrac{ 4pi^2a^3}{(6.67 imes 10^{-11} ext{m}^3)(6.04 imes 10^{24})} (5.563 imes 10^{12})(6.67 imes 10^{-11} ext{m}^3)(6.04 imes 10^{24}) = 4pi^2 a^3 a^3 = dfrac{2.241 imes 10^{27}}{4pi^2} ext{m}^3 a = 3.84 imes 10^8 ext{m} approx 384,000 , ext{km}. end{align*}]

Analysis

According to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is 384,400 km. This is calculated using reflectors left on the Moon by Apollo astronauts back in the 1960s.

Exercise (PageIndex{4})

Titan is the largest moon of Saturn. The mass of Titan is approximately (1.35 imes 10^{23} kg). The mass of Saturn is approximately ( 5.68 imes 10^{26}) kg. Titan takes approximately 16 days to orbit Saturn. Use this information, along with the universal gravitation constant (G=6.67×10^{−11} ext{m}/ ext{kg} cdot ext{sec}^2) to estimate the distance from Titan to Saturn.

Hint

Make sure your units agree, then use Equation ef{Eq30}.

Answer

[aapprox 1.224 imes 10^9 ext{m}= 1,224,000 ext{km} ]

Example (PageIndex{5}): Halley’s Comet

We now return to the chapter opener, which discusses the motion of Halley’s comet around the Sun. Kepler’s first law states that Halley’s comet follows an elliptical path around the Sun, with the Sun as one focus of the ellipse. The period of Halley’s comet is approximately 76.1 years, depending on how closely it passes by Jupiter and Saturn as it passes through the outer solar system. Let’s use (T=76.1) years. What is the average distance of Halley’s comet from the Sun?

Solution

Using the equation (T^2=D^3) with (T=76.1), we obtain (D^3=5791.21), so (Dapprox 17.96) A.U. This comes out to approximately (1.67 imes 10^9) mi.

A natural question to ask is: What are the maximum (aphelion) and minimum (perihelion) distances from Halley’s Comet to the Sun? The eccentricity of the orbit of Halley’s Comet is 0.967 (Source: http://nssdc.gsfc.nasa.gov/planetary...cometfact.html). Recall that the formula for the eccentricity of an ellipse is (e=c/a), where a is the length of the semimajor axis and c is the distance from the center to either focus. Therefore, (0.967=c/17.96) and (capprox 17.37) A.U. Subtracting this from a gives the perihelion distance (p=a−c=17.96−17.37=0.59) A.U. According to the National Space Science Data Center (Source: http://nssdc.gsfc.nasa.gov/planetary...cometfact.html), the perihelion distance for Halley’s comet is 0.587 A.U. To calculate the aphelion distance, we add

[ P=a+c=17.96+17.37=35.33 ; ext{A.U.} onumber ]

This is approximately (3.3 imes 10^9) mi. The average distance from Pluto to the Sun is 39.5 A.U. (Source: http://www.oarval.org/furthest.htm), so it would appear that Halley’s Comet stays just within the orbit of Pluto.

NAVIGATING A BANKED TURN

How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors:

  • The weight of the car;
  • The friction between the tires and the road;
  • The radius of the circle;
  • The “steepness” of the turn.

In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.

A car of mass (m) moves with constant angular speed (omega) around a circular curve of radius (R) (Figure (PageIndex{9})). The curve is banked at an angle ( heta). If the height of the car off the ground is (h), then the position of the car at time (t) is given by the function (vecs r(t)=< Rcos(omega t),Rsin(omega t),h>).

  1. Find the velocity function (vecs{v}(t)) of the car. Show that (vecs{v}) is tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.
  2. Show that the speed of the car is (omega R). Use this to show that ((2pi 4)/|vecs{v}|=(2pi)/omega ).
  3. Find the acceleration (vecs{a}). Show that this vector points toward the center of the circle and that (|vecs{a}|=Romega ^2).
  4. The force required to produce this circular motion is called the centripetal force, and it is denoted ( vecs{F}_{cent} ). This force points toward the center of the circle (not toward the ground). Show that (|vecs{F}_{cent}|=left(m|vecs{v}|^2 ight)/R).

As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure (PageIndex{10})). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that (vecs{f}=mu vecs{N}) for some positive constant (mu ). The constant (mu) is called the coefficient of friction.

Let (v_{max}) denote the maximum speed the car can attain through the curve without skidding. In other words, (v_{max}) is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is

[| vecs{F}_{cent} | = dfrac{m(v_{max})^2}{R}. ]

The next three questions deal with developing a formula that relates the speed (v_{max}) to the banking angle ( heta).

  1. Show that (vecs{N} cos heta=mvecs g+vecs{f} sin heta). Conclude that (vecs{N}=(mvecs g)/(cos heta−mu sin heta)).
  2. The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that

    [vecs{F}_{cent}=vecs{N} sin heta+vecs{f}cos heta.]

    Conclude that

    [vecs{F}_{cent}=dfrac{sin heta+mu cos heta}{cos heta−mu sin heta} mvecs g.]

  3. Show that ((v_{ ext{max}})^2=((sin heta+mu cos heta)/(cos heta−mu sin heta))gR). Conclude that the maximum speed does not actually depend on the mass of the car.
    Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.
    The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure (PageIndex{11}). Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately 211 ft with a banking angle of 24°. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of 28°. (The track has variable angle banking.)

The coefficient of friction for a normal tire in dry conditions is approximately 0.7. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately 0.98.

Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.

  1. In dry conditions, how fast can the car travel through the bottom of the turn without skidding?
  2. In dry conditions, how fast can the car travel through the top of the turn without skidding?
  3. In wet conditions, the coefficient of friction can become as low as 0.1. If this is the case, how fast can the car travel through the bottom of the turn without skidding?
  4. Suppose the measured speed of a car going along the outside edge of the turn is 105 mph. Estimate the coefficient of friction for the car’s tires.

Key Concepts

  • If (vecs{r}(t)) represents the position of an object at time t, then (vecs{r}'(t)) represents the velocity and (vecs{r}′′(t)) represents the acceleration of the object at time t. The magnitude of the velocity vector is speed.
  • The acceleration vector always points toward the concave side of the curve defined by (vecs{r}(t)). The tangential and normal components of acceleration (a_vecs{T}) and (a_vecs{N}) are the projections of the acceleration vector onto the unit tangent and unit normal vectors to the curve.
  • Kepler’s three laws of planetary motion describe the motion of objects in orbit around the Sun. His third law can be modified to describe motion of objects in orbit around other celestial objects as well.
  • Newton was able to use his law of universal gravitation in conjunction with his second law of motion and calculus to prove Kepler’s three laws.

Key Equations

  • Velocity [vecs{v}(t)=vecs{r}′(t) onumber]
  • Acceleration [vecs{a}(t)=vecs{v}′(t)=vecs{r}′′(t) onumber]
  • Speed [v(t)=||vecs{v}(t)||=||vecs{r}′(t)||=dfrac{ds}{dt} onumber]
  • Tangential component of acceleration [a_{vecs{T}} =vecs{a}cdot vecs{T}=dfrac{vecs{v}cdot vecs{a}}{||vecs v||} onumber]
  • Normal component of acceleration [a_{vecs{N}}=vecs{a}cdot vecs{N} = dfrac{|| vecs{v} imes vecs{a} ||}{||vecs{v}||} = sqrt{||vecs{a}||^2 - a_{vecs{T}}} onumber]

Glossary

acceleration vector
the second derivative of the position vector
Kepler’s laws of planetary motion
three laws governing the motion of planets, asteroids, and comets in orbit around the Sun
normal component of acceleration
the coefficient of the unit normal vector (vecs N) when the acceleration vector is written as a linear combination of (vecs T) and (vecs N)
projectile motion
motion of an object with an initial velocity but no force acting on it other than gravity
tangential component of acceleration
the coefficient of the unit tangent vector (vecs T) when the acceleration vector is written as a linear combination of (vecs T) and (vecs N)
velocity vector
the derivative of the position vector

Contributors and Attributions

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

  • Edited by Paul Seeburger
    Paul Seeburger added finding point ((1, 2)) when (t=1) and added part 3 (finding the equations of the tangent line) in Example (PageIndex{1}).
    He also created Figure (PageIndex{1}).

1.त्वरण के स्पर्शरेखीय तथा अभिलाम्बिक घटक (Tangential and normal components of acceleration,Tangential and normal components of velocity and acceleration)-

त्वरण के स्पर्शरेखीय तथा अभिलाम्बिक घटक (Tangential and normal components of acceleration),स्पर्शरेखीय तथा अभिलाम्बिक वेग पर निर्भर करता है।
स्पर्शरेखीय त्वरण वेग वेक्टर के परिमाण में परिवर्तन की दर का माप है, अर्थात गति और अभिलाम्बिक त्वरण वेग वेक्टर की दिशा के परिवर्तन की दर का एक माप है।

आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके ।यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए ।आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं। इस आर्टिकल को पूरा पढ़ें।

2.आप स्पर्शरेखीय और अभिलाम्बिक त्वरण कैसे ज्ञात करते हैं? (How do you find tangential and normal acceleration?)-(1.)स्पर्शरेखीय तथा अभिलाम्बिक वेग (Tangential and Normal velocities)-

मान लो वक्र पर एक स्थिर बिन्दु A है तथा इस पर चलता हुआ कोई कण t समय पर P बिन्दु पर पहुंचता है, जहां चाप AP=s।मानलो कि कण t+delta t समय पर बिन्दु Q पर है, जहां चाप PQ= delta s
यह भी मानलो कि वक्र के बिन्दु P पर स्पर्श रेखा x-अक्ष के साथ psi कोण बनाती है।यदि Q से,P बिन्दु पर खींची गई स्पर्शरेखा पर QM लम्ब डालें तब कण का समय में P पर स्पर्शरेखा के अनुदिश विस्थापन PM होगा।

अतः P पर स्पर्शरेखीय वेग (Tangential Velocity)
= egin lim delta t ightarrow 0 endfrac < delta t समय quad में P पर quad स्पर्शरेखा quad के quad अनुदिश quad विस्थापन> < delta t > =egin lim delta t ightarrow 0 endfrac < PM > < delta t > =egin lim delta t ightarrow 0 endfrac < जीवाPQquad cos < QPM >> < delta t > =egin lim delta t ightarrow 0 endfrac < जीवाPQquad > < delta s >.frac < delta s > < delta t >cos < QPM > =1.frac < delta s > < delta t >.1
[ ecause जब quad Q ightarrow P तब frac <जीवा pq=""> < delta s >=1 तथा angle QPM ightarrow 0 ]
P पर स्पर्शरेखीय वेग= frac < ds > < dt >=overset < ullet > < s >
अब यदि PN ,P पर अन्दर की ओर अभिलम्ब (अवतल दिशा की ओर) हो,तब delta t समय में PN के समान्तर विस्थापन MQ होगा।
अतः P पर अभिलाम्बिक वेग (Normal Velocity)
= lim _< delta t ightarrow 0 >< frac < MOquad > < delta t >> =lim _< delta t ightarrow 0 > < frac < जीवाPQsin < QPM >quad > < delta t >> =lim _< delta t ightarrow 0 >< frac < जीवाPQquad > < delta s >> .frac < delta squad > < delta t >sin < QPM > =1.frac < delta squad > < delta t >.0
[ ecause जबquad Q ightarrow P ,तब frac < जीवा PQquad > < delta s >=1 तथा angle QPM ightarrow 0 ]
P पर अभिलाम्बिक वेग=0
अतः एक कण जो समतल में किसी वक्र के अनुदिश गमन करता है,का स्पर्शरेखीय वेग frac < ds > < dt >तथा अभिलाम्बिक वेग 0 होता है।

(2.)त्वरण के स्पर्शरेखीय तथा अभिलाम्बिक घटक (Tangential and normal components of acceleration,Tangential and normal components of velocity and acceleration),त्वरण के स्पर्शरेखीय घटक का सूत्र (Tangential component of acceleration formula)-

मान लो t व t+delta t समय पर कण की स्थिति बिन्दु P व Q पर है।मानलो P व Q पर कण का वेग क्रमशः V तथा t+delta V है।यह भी मानलो कि P व Q पर खींची PT व QT' x-अक्ष के साथ क्रमशः कोण psi व psi +delta psi बनाती है तब स्पर्श रेखाओं के मध्य का कोण होगा।

अतः समय अन्तराल में TP के अनुदिश वेग में परिवर्तन
=(Q पर TP के अनुदिश वेग)-(P पर TP के अनुदिश वेग)

= left( v+delta v ight) cos < delta psi >-v =left( v+delta v ight) .1-v =delta v [ delta psi छोटा है, herefore cos < delta psi > ightarrow 1 ]
इसी प्रकार समय अन्तराल में TP के लम्बवत वेग में परिवर्तन
=Q पर TP के लम्बवत वेग-P पर TP के लम्बवत वेग

= left( v+delta v ight) sin < delta psi >-0 =left( v+delta v ight) delta psi =vdelta psi [ delta psi छोटा है, herefore sin < delta psi > ightarrow delta psi ,delta vdelta psi ightarrow 0 ]
अतः P पर स्पर्शरेखीय त्वरण (Tangential Acceleration)
=lim _< delta t ightarrow 0 >< frac < delta t समय quad अन्तराल quad में P के quad अनुदिश quad वेग quad में quad परिवर्तन> < delta t >> =lim _< delta t ightarrow 0 >< frac < delta v > < delta t >> =frac < dv > < dt >=frac < d > < dt >left( frac < ds > < dt > ight) =frac < < d >^< 2 >s >< d< t >^ < 2 >> =frac < dv > < ds >.frac < ds > < dt >=vfrac < dv > < ds >
अतः स्पर्शरेखीय त्वरण= frac < dv > < dt >=frac < < d >^< 2 >s >< d< t >^ < 2 >> =vfrac < dv > < ds >
तथा P पर अभिलाम्बिक त्वरण (Normal Acceleration)
= lim _< delta t ightarrow 0 >< frac < delta t समय quad अन्तराल quad में TP के quad लम्बवत quad वेग quad में quad परिवर्तन> < delta t >> lim _< delta t ightarrow 0 >< vfrac < delta psi > < delta t >> =vfrac < dpsi > < dt > =vfrac < dpsi > < ds >.frac < ds > < dt >=v.frac < 1 > < ho >.v=frac < < v >^ < 2 >> < ho >
[ ecause ho =frac < ds > < dpsi >जहां ho वक्र के बिन्दु P पर वक्रता त्रिज्या]
अतः अभिलाम्बिक त्वरण= frac < < v >^ < 2 >> < ho >
टिप्पणी:स्पर्शरेखीय वेग तथा त्वरण की धनात्मक दिशा के बढ़ने की दिशा में होगी तथा अभिलाम्बिक त्वरण अभिलम्ब के अनुदिश अन्दर की तरफ होगा।
विशेष स्थिति (Particular Case): यदि कण एक वृत्तीय पथ (Circular Path) पर चले तो s=a heta तथा ho =a
स्पर्शरेखीय वेग frac < ds > < dt >=adot < heta >अनुप्रस्थ वेग

अभिलाम्बिक त्वरण=0=अरीय वेग
स्पर्शरेखीय त्वरण= frac < < d >^< 2 >s >< d< t >^ < 2 >> =addot < heta >अनुप्रस्थ त्वरण
अभिलाम्बिक त्वरण= frac < < v >^ < 2 >> < ho >=frac < < left( adot < heta > ight) >^ < 2 >> < ho >=a < dot < heta >>^ < 2 >अरीय त्वरण


3 Answers 3

If the speed is constant there is only normal acceleration (from the motorist to the center of the circunference).

To get both accelerations you can make:

1) Get your position vector in polar coordinates: $ vec = xvec+yvec =r cos heta hat + r sin heta hat = r hat. $

Now you can get the speed vector making the derivative with respect to time: $ vec = dfrac<>>

=omega r left( - sin heta hat + cos heta hat ight) = omega r hat, $ wehre $ omega = dfrac
. $

If you go back to the original coordinates, you get what you said: $ a_n = dfrac, qquad a_t = dfrac

. $ In your homework notice that you hace a 70 km/h speed with no acceleration. Then one of the circular accelerations is zero. Guess what? Notice also that in a circular motion theres is ALWAYS one acceleration that MUST be non-zero.


Calculate tangential and normal components of acceleration

Q: Given a particule that moves with a position function: r(t) = <cos(2t),1,-sin(2t)>,
d) Calculate tangential and normal components of acceleration when t=pi/8
e) Is the particle speeding up or slowing down , and turning or going straight when t=pi/8. Why?

This is a long multipart problem, so I just posted two parts. These are the two parts that I don't have a clue how to tackle.
I can get the acceleration for this particle when pi/8 by taking the second derivative r''(t), but i'm not sure if that corresponds to either the tangential or normal, or how to distinguish between the two. I am also clueless about how to solve part e.

I'm thinking this has to do with the unit tangent vector, principle unit normal vector. I'm not sure how to relate those to this problem though.

Subhotosh Khan

Super Moderator

Q: Given a particule that moves with a position function: r(t) = <cos(2t),1,-sin(2t)>,
d) Calculate tangential and normal components of acceleration when t=pi/8
e) Is the particle speeding up or slowing down , and turning or going straight when t=pi/8. Why?

Normal (radial) unit vector

(displaystyle e_r , = , frac<1>> cdot [cos 2t, , 1, , -sin 2t])

If you take a cross product of this vector with acceleration vector - you'll get the tangential component of the acceleration.

ofcourse dot product will give you the normal component.


This is a long multipart problem, so I just posted two parts. These are the two parts that I don't have a clue how to tackle.
I can get the acceleration for this particle when pi/8 by taking the second derivative r''(t), but i'm not sure if that corresponds to either the tangential or normal, or how to distinguish between the two. I am also clueless about how to solve part e.

I'm thinking this has to do with the unit tangent vector, principle unit normal vector. I'm not sure how to relate those to this problem though.


[Vector Calculus] Trying to understand formula and geometric representation of Tangential and Normal Component for Acceleration

I am trying to understand the formula for the tangential and normal components of acceleration geometrically.

*aTAN = tangential component for acceleration

*aNORM = normal component for acceleration

*T = tangential unit vector

So to start, the formula for the acceleration vector is

a = (aTAN)(T) + (aNORM)(N), which means the vector a lies in the plane of the unit vectors T and N.

So the formula the textbook gives me for aTAN is

aTAN = d/dt[||v||] = aT = va / ||v||

so obviously using the definition of a dot product

va = ||v|| ||a|| cos(θ)

(divide by ||v||)

which makes va / ||v|| = ||a|| cos(θ)

Well In this IMAGE you can see that aNORM = ||a|| cos(θ), but in my formula aTAN = ||a|| cos(θ)

My question is, how come the image shows aNORM = ||a|| cos(θ) instead of aTAN = ||a|| cos(θ), same thing can be said for the other way around.

Another way to consider this is by thinking of the tangential and normal components of the acceleration as parallel and orthogonal projections of the acceleration onto the velocity. The parallel part (dot product) is what speeds up or slows down the object (increases or decreases the magnitude of the velocity), and the orthogonal part (cross product) is what changes the direction/turns the object.

As the other poster said you are using two different angles. To be consistent with the dot product formula, you should call the angle between the unit tangent (the unit vector pointing in the direction of the velocity vector) and the acceleration theta in the dot product, not the angle between the normal (the normal is orthogonal to the velocity) and the acceleration.

This is now making sense to me. I made the wrong assumption of labeling the normal vector as horizontal since it looks horizontal.

So taking what you said the previous image should look like this.

On top of that, my guess is I would use θ1 for the formulas.

EDIT: OKAY I understand, the dot product can give you the angel between two vectors, so we're not even looking at x and y axis, we're just finding the angle between the acceleration vector, a, and the tangent unit vector, T.

You're using two different θs because all three vectors are coplanar and the tangent and normal vectors are perpendicular, the angles between a and its tangent and normal components are complementary.

The way Iɽ do it, the tangent is more "horizontal-like", so Iɽ use θ as the angle between a and its tangent component then π/2-θ is the angle between a and its normal component, and cos(π/2-θ)=sin(θ), so the magnitude of the normal component is |a|sin(θ).

Okay I understand now, because when I look at other images such as this one the tangential component is undoubtedly horizontal.

So in my case I should use the angel between the tangent unit vector and acceleration vector like this.

I do have one more question though, would I use θ1 or θ2? My assumption is that the tangent unit vector is pointing in the +x direction (upward), instead of the -x direction (downward). As a result, I would use θ1. This is hard for me to grasp since I've always dealt with +x direction is to the right and -x direction to the left.

I really appreciate your help by the way. it seems like I'm making things way more difficult for me than it should be, but at the same time I'm just trying to understand the concept. Thanks

EDIT: I get it now, instead of looking at the tangent vector as an x axis and the normal vector as a y axis, we're looking at θ as the angle between the two vectors!


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Is acceleration the rate of change of speed?

Is this true or false?

Acceleration is the rate of change of speed.

Why some people say it's true: Think of accelerating in a car: when you hit the gas, you speed up, and when you hit the brake, you slow down. Acceleration is generally associated with a change in speed.

Why some people say it's false: In physics, direction matters. If the direction of motion changes, this could be considered acceleration too, even if the speed stays constant.

Explanation:

Acceleration is defined as the rate of change of velocity. Velocity is a vector, which means it contains a magnitude (a numerical value) and a direction. So the velocity can be changed either by changing the speed or by changing the direction of motion (or both). Therefore, it may be possible that the speed is constant, but the velocity is changing because the direction is changing. In this case, acceleration will be non-zero and equal to the rate of change of velocity.

It is a general misconception that rate of change of speed is equal to the magnitude of the rate of change of velocity. However, this is not true in all cases. Consider uniform circular motion: in the case of a uniform circular motion, the particle moves on a circular path with uniform speed. The speed remains constant, but the direction of motion is continuously changing. Due to change in direction of motion, acceleration is non-zero. This acceleration is toward the center of the circle and known as centripetal acceleration.

In general, acceleration can be resolved into two components. One component, which is parallel to the velocity, is known as tangential acceleration. This component changes the speed of the particle and is equal to the rate of change of speed. The other component of acceleration, which is perpendicular to velocity, is known as normal acceleration. This component is responsible for changing the direction of the velocity.

Velocity is the rate of change of displacement, while speed is the rate of change of distance. In other words, velocity is the rate of change of the shortest distance moved by a body from the final position to the initial position, while speed is the rate of change of the total length of the path traveled by a certain body.

Query: What can be said about the acceleration of a particle moving in a zig-zag path with constant speed?
Reply: The acceleration of the particle has to be non-zero, as the particle is changing its direction. However, the tangential component of the acceleration is zero, as the speed remains constant.

Query: If both the speed and direction changes, then is it possible to have zero acceleration?
Reply: No. If speed changes, then tangential acceleration is non-zero. If the direction of motion is changing, then normal acceleration is non-zero. The result of these two accelerations can never be zero as they are perpendicular to each other.

A particle is moving on a circular track with constant non-zero speed. Which of the following options are correct?

(a) The acceleration of the particle is zero.

(b) The rate of change of speed equals the magnitude of the rate of change of velocity.

(c) Instantaneous speed equals the magnitude of instantaneous velocity.

(d) The angle between velocity and acceleration has to be 9 0 ∘ 90^circ 9 0 ∘ .


12.5: Tangential and Normal Components of Acceleration - Mathematics

The following are lectures for Calculus III - Multivariable. They are mostly in Adobe pdf format. Clicking on a link will take you to Google Docs.

Notes on Google Docs: You should be able do view these files on Google Docs or download them. To download, either click on "Download" if you see that option, or click on "File" and then "Download Original" if you see that option. Unfortunately, sometimes Google Docs does not work with the Internet Explorer browser. If you are unable to download the files using Internet Explorer, you may need to install the Google Chrome browser. To download the Google Chrome browser, click HERE.

A few of these lectures are PowerPoint presentations. To view them on an iPad, download the free SlideShark application from iTunes.

Some of these lectures reference the TI-89 graphing calculator.

Lectures with an N after the lecture number have been rewritten to reference the TI-nspire graphing calculator.

The dates by some of the lectures are the date of the most recent revision.


Position, velocity and acceleration vectors

Kinematics is the branch of Mechanics that describes the motion of particles, objects or groups of objects. In these pages we will analyse the motion of a point particle in any trajectory like for instance the one shown in black in the figure below. For the sake of simplicity, a flat trajectory has been represented in the figure, but the trajectory can be three-dimensional.

A reference frame at rest and an observer O located at the origin of a Cartesian coordinate system are also shown in the previous figure. This reference frame is called inertial. The orientation of the three Cartesian axes is indicated by the unit vectors i, j and k respectively. We will describe the motion of the particle with respect to this reference frame.

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The motion of a particle is described by three vectors: position, velocity and acceleration.

The position vector (represented in green in the figure) goes from the origin of the reference frame to the position of the particle. The Cartesian components of this vector are given by:

The components of the position vector are time dependent since the particle is in motion. In order to simplify the notation we will often omit this dependence in the expressions of the vectors.

The velocity vector is the time derivative of the position vector:

Which can also be expressed as:

The velocity vector is always tangent to the trajectory of the particle at each point.

The acceleration vector is the time derivative of the velocity vector:

Which can also be expressed as:

The acceleration vector is the variation of the velocity vector over time. Therefore, it must always be directed towards the inside of the particle trajectory, as shown in the figure.

In these pages you will find numerous problems where you will learn to calculate these three vectors in different situations.

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The acceleration vector can be expressed as a function of its projections on a reference frame moving with the particle and with axes that are respectively tangent and perpendicular (or normal) for each point of its trajectory. These projections are called tangential acceleration and normal acceleration (or centripetal).

The previous figure represents the acceleration vector expressed as the sum of these components.

The tangential acceleration is given by:

Where ut is a unit vector tangent to the path at each point that is determined by dividing the velocity vector by its magnitude:

The tangential acceleration provides information about the variation of the velocity vector magnitude.

On the other hand, the normal acceleration (or centripetal) is given by:

Where un is a unit vector perpendicular to the path at each point and ρ is the radius of curvature of the path.

The normal acceleration provides information about the variation of the velocity vector’s direction. If a particle describes a straight line, the radius of curvature is infinite and therefore its normal acceleration is zero.

In the particular case of a circular path, the normal acceleration magnitude is:

The different problems you will find in these pages will help you to learn how to calculate the components of the acceleration vector.


12.5: Tangential and Normal Components of Acceleration - Mathematics

Consider an object which appears stationary in our rotating reference frame: i.e. , an object which is stationary with respect to the Earth's surface. According to Equation (414), the object's apparent equation of motion in the rotating frame takes the form

Let the non-fictitious force acting on our object be the force of gravity, . Here, the local gravitational acceleration, , points directly toward the center of the Earth. It follows, from the above, that the apparent gravitational acceleration in the rotating frame is written

where is the displacement vector of the origin of the rotating frame (which lies on the Earth's surface) with respect to the center of the Earth. Here, we are assuming that our object is situated relatively close to the Earth's surface ( i.e. , ).

It can be seen, from Equation (417), that the apparent gravitational acceleration of a stationary object close to the Earth's surface has two components. First, the true gravitational acceleration, , of magnitude , which always points directly toward the center of the Earth. Second, the so-called centrifugal acceleration , . This acceleration is normal to the Earth's axis of rotation, and always points directly away from this axis. The magnitude of the centrifugal acceleration is , where is the perpendicular distance to the Earth's rotation axis, and is the Earth's radius--see Figure 25.

It is convenient to define Cartesian axes in the rotating reference frame such that the -axis points vertically upward, and - and -axes are horizontal, with the -axis pointing directly northward, and the -axis pointing directly westward--see Figure 24. The Cartesian components of the Earth's angular velocity are thus

whilst the vectors and are written

respectively. It follows that the Cartesian coordinates of the apparent gravitational acceleration, (417), are

The magnitude of this acceleration is approximately

According to the above equation, the centrifugal acceleration causes the magnitude of the apparent gravitational acceleration on the Earth's surface to vary by about , being largest at the poles, and smallest at the equator. This variation in apparent gravitational acceleration, due (ultimately) to the Earth's rotation, causes the Earth itself to bulge slightly at the equator (see Section 12.6), which has the effect of further intensifying the variation, since a point on the surface of the Earth at the equator is slightly further away from the Earth's center than a similar point at one of the poles (and, hence, the true gravitational acceleration is slightly weaker in the former case).

Another consequence of centrifugal acceleration is that the apparent gravitational acceleration on the Earth's surface has a horizontal component aligned in the north/south direction. This horizontal component ensures that the apparent gravitational acceleration does not point directly toward the center of the Earth. In other words, a plumb-line on the surface of the Earth does not point vertically downward, but is deflected slightly away from a true vertical in the north/south direction. The angular deviation from true vertical can easily be calculated from Equation (421):