Kinematics of Rotational Motion

Kinematics of Rotational Motion

Just by using our intuition, we can begin to see how rotational quantities like \theta, \omega, and \alpha are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel's angular acceleration \alpha is large for a long period of time t, then the final angular velocity \omega and angle of rotation \theta are large. The wheel's rotational motion is exactly analogous to the fact that the motorcycle's large translational acceleration produces a large final velocity, and the distance traveled will also be large.

Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating \omega, \alpha, and t. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:

 v=v_{0}+a t \quad(\operatorname{constant} a)

Note that in rotational motion a=a_{\mathrm{t}}, and we shall use the symbol a for tangential or linear acceleration from now on. As in linear kinematics, we assume a is constant, which means that angular acceleration \alpha is also a constant, because a=r \alpha. Now, let us substitute v=r \omega and a=r \alpha into the linear equation above:

 r \omega=r \omega_{0}+r \alpha t.

The radius r cancels in the equation, yielding

 \omega=\omega_{0}+\alpha t \quad(\operatorname{constant} \alpha),

where \omega_{0} is the initial angular velocity. This last equation is a kinematic relationship among \omega, \alpha, and t- that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart.

Making Connections

Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion.

Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts):

Rotational Translational  
\theta=\bar{\omega} t x=\bar{v} t  
\omega=\omega_{0}+\alpha t v=v_{0}+a t (constant \alpha, a))
\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2} x=v_{0} t+\frac{1}{2} a t^{2} (constant \alpha, a))
\omega^{2}=\omega_{0}^{2}+2 \alpha \theta v^{2}=v_{0}^{2}+2 a x (constant \alpha, a)

Table 10.2
Rotational Kinematic Equations

In these equations, the subscript 0 denotes initial values (  \theta_{0}, x_{0} , and t_{0} are initial values), and the average angular velocity \bar{\omega} and average velocity \bar{v} are defined as follows:

\bar{\omega}=\frac{\omega_{0}+\omega}{2} \text { and } \bar{v}=\frac{v_{0}+v}{2}.

The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which a and \alpha are constant.

Problem-Solving Strategy for Rotational Kinematics

  1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion.

  2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.

  3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).

  4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion.

  5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles.

  6. Check your answer to see if it is reasonable: Does your answer make sense?

Example 10.3 Calculating the Acceleration of a Fishing Reel

A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 \, \mathrm{rad} / \mathrm{s}^{2}+ for 2.00 s as seen in Figure 10.8.

(a) What is the final angular velocity of the reel?

(b) At what speed is fishing line leaving the reel after 2.00 s elapses?

(c) How many revolutions does the reel make?

(d) How many meters of fishing line come off the reel in this time?


In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown.

Solution for (a)

Here \alpha and t are given and \omega needs to be determined. The most straightforward equation to use is \omega=\omega_{0}+\alpha t because the unknown is already on one side and all other terms are known. That equation states that

 \omega=\omega_{0}+\alpha t.

We are also given that \omega_{0}=0 (it starts from rest), so that

\omega=0+\left(110 \, \mathrm{rad} / \mathrm{s}^{2}\right)(2.00 \mathrm{~s})=220 \, \mathrm{rad} / \mathrm{s}.

Solution for (b)

Now that \omega is known, the speed v can most easily be found using the relationship

v=r \omega,

where the radius r of the reel is given to be 4.50 \mathrm{~cm}; thus,

 v=(0.0450 \mathrm{~m})(220 \, \mathrm{rad} / \mathrm{s})=9.90 \mathrm{~m} / \mathrm{s}.

Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have \mathrm{m} \times \mathrm{rad}=\mathrm{m}.

Solution for (c)

Here, we are asked to find the number of revolutions. Because 1 \mathrm{rev}=2 \pi \mathrm{rad}, we can find the number of revolutions by finding \theta in radians. We are given \alpha and t, and we know \omega_{0} is zero, so that \theta can be obtained using \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}.

\begin{aligned} \theta &=\omega_{0} t+\frac{1}{2} \alpha t^{2} \\ &=0+(0.500)\left(110 \, \mathrm{rad} / \mathrm{s}^{2}\right)(2.00 \mathrm{~s})^{2}=220 \, \mathrm{rad}.

Converting radians to revolutions gives

\theta=(220 \, \mathrm{rad}) \frac{1 \, \mathrm{rev}}{2 \pi \, \mathrm{rad}}=35.0 \, \mathrm{rev}.

Solution for (d)

The number of meters of fishing line is x, which can be obtained through its relationship with \theta:

x=r \theta=(0.0450 \mathrm{~m})(220 \, \mathrm{rad})=9.90 \mathrm{~m}.


This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds). The amount of fishing line played out is 9.90 m, about right for when the big fish bites.

The figure shows a fishing reel, with radius equal to 4.5 centimeters. The direction of rotation of the reel is counterclockwise. The rotational quantities are theta, omega and alpha, and x, v, a are linear or translational quantities. The reel, fishing line, and the direction of motion have been separately indicated by curved arrows pointing toward those parts.

Figure 10.8 Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel.

Example 10.4 Calculating the Duration When the Fishing Reel Slows Down and Stops

Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of -300 \mathrm{rad} / \mathrm{s}^{2}. How long does it take the reel to come to a stop?


We are asked to find the time t for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is \omega_{0}=220
            \mathrm{rad} / \mathrm{s} and the final angular velocity \omega is zero. The angular acceleration is given to be \alpha=-300 \mathrm{rad} / \mathrm{s}^{2}. Examining the available equations, we see all quantities but t are known in \omega=\omega_{0}+\alpha t, making it easiest to use this equation.


The equation states 

 \omega=\omega_{0}+\alpha t.

We solve the equation algebraically for t, and then substitute the known values as usual, yielding

t=\frac{\omega-\omega_{0}}{\alpha}=\frac{0-220 \, \mathrm{rad} / \mathrm{s}}{-300 \, \mathrm{rad} / \mathrm{s}^{2}}=0.733 \mathrm{~s}.


Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration.

Example 10.5 Calculating the Slow Acceleration of Trains and Their Wheels

Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350 m-radius wheels an angular acceleration of 0.250 \, \mathrm{rad} / \mathrm{s}^{2}. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train?


In part (a), we are asked to find x, and in (b) we are asked to find \omega and v. We are given the number of revolutions \theta, the radius of the wheels r, and the angular acceleration \alpha.

Solution for (a)

The distance x is very easily found from the relationship between distance and rotation angle:


Solving this equation for x yields

x=r \theta.

Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities:

\theta=(200 \text { rev }) \frac{2 \pi \, \mathrm{rad}}{1 \, \mathrm{rev}}=1257 \, \mathrm{rad}

Now we can substitute the known values into x=r \theta to find the distance the train moved down the track:

x=r \theta=(0.350 \mathrm{~m})(1257 \, \mathrm{rad})=440 \mathrm{~m}.

Solution for (b)

We cannot use any equation that incorporates t to find \omega, because the equation would have at least two unknown values. The equation \omega^{2}=\omega_{0}^{2}+2 \alpha \theta will work, because we know the values for all variables except \omega:

\omega^{2}=\omega_{0}^{2}+2 \alpha \theta.

Taking the square root of this equation and entering the known values gives

\begin{aligned}\omega &=\left[0+2\left(0.250 \mathrm{rad} / \mathrm{s}^{2}\right)(1257 \mathrm{rad})\right]^{1 / 2} \\ &=25.1 \mathrm{rad} / \mathrm{s}.\end{aligned}

We can find the linear velocity of the train, v, through its relationship to \omega :

v=r \omega=(0.350 \mathrm{~m})(25.1 \mathrm{rad} / \mathrm{s})=8.77 \mathrm{~m} / \mathrm{s}.

The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h).

There is translational motion even for something spinning in place, as the following example illustrates. Figure 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels.

The figure shows a fly that has landed on the rotating plate of the microwave. The direction of rotation of the plate, omega, is counterclockwise and is shown with an arrow.

Figure 10.9 The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly).

Example 10.6 Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate

A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate, and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times).


First, find the total number of revolutions \theta, and then the linear distance x traveled. \theta=\bar{\omega} t can be used to find \theta because \bar{\omega} is given to be 6.0 rpm.


Entering known values into \theta=\bar{\omega} t gives

\theta=\bar{\omega} t=(6.0 \, \mathrm{rpm})(2.0 \, \mathrm{~min})=12 \, \mathrm{rev}.

As always, it is necessary to convert revolutions to radians before calculating a linear quantity like x from an angular quantity like \theta:

 \theta=(12 \text { rev })\left(\frac{2 \pi \, \mathrm{rad}}{1 \, \mathrm{rev}}\right)=75.4 \, \mathrm{rad}.

Now, using the relationship between x and \theta, we can determine the distance traveled:

x=r \theta=(0.15 \mathrm{~m})(75.4 \, \mathrm{rad})=11 \mathrm{~m}.


Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics.


Source: Rice University,
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