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College Physics 2e

10.2 Kinematics of Rotational Motion

College Physics 2e10.2 Kinematics of Rotational Motion

Learning Objectives

By the end of this section, you will be able to:

  • Observe the kinematics of rotational motion.
  • Derive rotational kinematic equations.
  • Evaluate problem solving strategies for rotational kinematics.

Just by using our intuition, we can begin to see how rotational quantities like θθ, ωω, and αα 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 αα is large for a long period of time tt, then the final angular velocity ωω and angle of rotation θθ 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 ωω, αα, and tt. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:

v = v 0 + at       ( constant  a ) v = v 0 + at       ( constant  a )

Note that in rotational motion a=ata=at, and we shall use the symbol aa for tangential or linear acceleration from now on. As in linear kinematics, we assume aa is constant, which means that angular acceleration αα is also a constant, because a=a=. Now, let us substitute v=v= and a=a= into the linear equation above:

= 0 + rαt . = 0 + rαt .

The radius rr cancels in the equation, yielding

ω = ω 0 + αt       ( constant  α ) , ω = ω 0 + αt       ( constant  α ) ,

where ω0ω0 is the initial angular velocity. This last equation is a kinematic relationship among ωω, αα, and tt —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
θ = ω ¯ t θ = ω ¯ t x = v - t x = v - t
ω = ω 0 + αt ω = ω 0 + αt v = v 0 + at v = v 0 + at (constant αα, aa)
θ = ω 0 t + 1 2 αt 2 θ = ω 0 t + 1 2 αt 2 x = v 0 t + 1 2 at 2 x = v 0 t + 1 2 at 2 (constant αα, aa)
ω 2 = ω 0 2 + 2 αθ ω 2 = ω 0 2 + 2 αθ v 2 = v 0 2 + 2 ax v 2 = v 0 2 + 2 ax (constant αα, aa)
Table 10.2 Rotational Kinematic Equations

In these equations, the subscript 0 denotes initial values (θ0θ0, x0x0, and t0t0 are initial values), and the average angular velocity ω-ω- and average velocity v-v- are defined as follows:

ω ¯ = ω 0 + ω 2  and  v ¯ = v 0 + v 2 . ω ¯ = ω 0 + ω 2  and  v ¯ = 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 aa and αα 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 110rad/s2110rad/s2 for 2.00 s as seen in Figure 10.7.

(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 αα and tt are given and ωω needs to be determined. The most straightforward equation to use is ω=ω0+αtω=ω0+αt because the unknown is already on one side and all other terms are known. That equation states that

ω=ω0+αt .ω=ω0+αt .

We are also given that ω0=0ω0=0 (it starts from rest), so that

ω=0+110 rad/s22.00 s=220rad/s .ω=0+110 rad/s22.00 s=220rad/s .

Solution for (b)

Now that ωω is known, the speed vv can most easily be found using the relationship

v= ,v= ,

where the radius rr of the reel is given to be 4.50 cm; thus,

v=0.0450 m220 rad/s=9.90 m/s.v=0.0450 m220 rad/s=9.90 m/s.

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

Solution for (c)

Here, we are asked to find the number of revolutions. Because 1 rev=2π rad1 rev=2π rad, we can find the number of revolutions by finding θθ in radians. We are given αα and tt, and we know ω0ω0 is zero, so that θθ can be obtained using θ=ω0t+12αt2θ=ω0t+12αt2.

θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad . θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad .

Converting radians to revolutions gives

θ=220 rad1 rev2π rad=35.0 rev.θ=220 rad1 rev2π rad=35.0 rev.

Solution for (d)

The number of meters of fishing line is xx, which can be obtained through its relationship with θθ:

x = = 0.0450 m 220 rad = 9.90 m . x = = 0.0450 m 220 rad = 9.90 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 10.7 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 300rad/s2300rad/s2. How long does it take the reel to come to a stop?


We are asked to find the time tt 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 ω0=220 rad/sω0=220 rad/s and the final angular velocity ωω is zero. The angular acceleration is given to be α=300rad/s2α=300rad/s2. Examining the available equations, we see all quantities but t are known in ω=ω0+αt,ω=ω0+αt, making it easiest to use this equation.


The equation states

ω=ω0+αt .ω=ω0+αt .

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

t=ωω0α=0220 rad/s300rad/s2=0.733 s.t=ωω0α=0220 rad/s300rad/s2=0.733 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 0.250rad/s20.250rad/s2. 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 xx, and in (b) we are asked to find ωω and vv. We are given the number of revolutions θθ, the radius of the wheels rr, and the angular acceleration αα.

Solution for (a)

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

θ = x r . θ = x r .

Solving this equation for xx yields


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:

θ = 200 rev rad 1 rev = 1257 rad . θ = 200 rev rad 1 rev = 1257 rad .

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

x = = 0.350 m 1257 rad = 440 m . x = = 0.350 m 1257 rad = 440 m .

Solution for (b)

We cannot use any equation that incorporates tt to find ωω, because the equation would have at least two unknown values. The equation ω2= ω02+2αθω2= ω02+2αθ will work, because we know the values for all variables except ωω:

ω2= ω02+2αθω2= ω02+2αθ

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

ω = 0 + 2 ( 0 . 250  rad/s 2 ) ( 1257  rad ) 1 / 2 = 25.1 rad/s. ω = 0 + 2 ( 0 . 250  rad/s 2 ) ( 1257  rad ) 1 / 2 = 25.1 rad/s.

We can find the linear velocity of the train, vv, through its relationship to ωω:

v = = 0.350 m 25.1 rad/s = 8.77 m/s . v = = 0.350 m 25.1 rad/s = 8.77 m/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 10.8 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 10.8 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 θθ, and then the linear distance xx traveled. θ=ω¯tθ=ω¯t can be used to find θθ because ω - ω - is given to be 6.0 rpm.


Entering known values into θ=ω¯tθ=ω¯t gives

θ=ω-t=6.0 rpm2.0 min=12 rev.θ=ω-t=6.0 rpm2.0 min=12 rev.

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

θ=12 rev2π rad1 rev=75.4 rad.θ=12 rev2π rad1 rev=75.4 rad.

Now, using the relationship between xx and θθ, we can determine the distance traveled:

x = = 0 . 15  m 75 . 4  rad = 11  m . x = = 0 . 15  m 75 . 4  rad = 11  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.

Check Your Understanding

Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.)

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