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=v0+at     (constant a)𝑣=𝑣0+at     (constant 𝑎)

10.17

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=rα𝑎=rα. Now, let us substitute v=rω𝑣=rω and a=rα𝑎=rα into the linear equation above:

rω=rω0+rαt.rω=rω0+rαt.

10.18

The radius r𝑟 cancels in the equation, yielding

ω=ω0+αt     (constant α),𝜔=𝜔0+αt     (constant 𝛼),

10.19

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¯¯t𝜃=𝜔¯𝑡	x=v–t𝑥=𝑣-𝑡
ω=ω0+αt𝜔=𝜔0+αt	v=v0+at𝑣=𝑣0+at	(constant α𝛼, a𝑎)
θ=ω0t+12αt2𝜃=𝜔0𝑡+12αt2	x=v0t+12at2𝑥=𝑣0𝑡+12at2	(constant α𝛼, a𝑎)
ω2=ω02+2αθ𝜔2=𝜔02+2αθ	v2=v02+2ax𝑣2=𝑣02+2ax	(constant α𝛼, a𝑎)

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¯¯=ω0+ω2 and v¯=v0+v2.𝜔¯=𝜔0+𝜔2 and 𝑣¯=𝑣0+𝑣2.

10.20

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?

Strategy

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.

10.21

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

ω=0+(110rad/s2)(2.00s)=220rad/s.𝜔=0+110rad/s22.00s=220rad/s.

10.22

Solution for (b)

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

v=rω,𝑣=rω,

10.23

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

v=(0.0450 m)(220 rad/s)=9.90 m/s.𝑣=0.0450 m220 rad/s=9.90 m/s.

10.24

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=mm×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==ω0t+12αt20+(0.500)(110rad/s2)(2.00 s)2=220 rad.𝜃=𝜔0𝑡+12αt2=0+0.500110rad/s22.00 s2=220 rad.

10.25

Converting radians to revolutions gives

θ=(220 rad)1 rev2π rad=35.0 rev.𝜃=220 rad1 rev2π rad=35.0 rev.

10.26

Solution for (d)

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

x=rθ=(0.0450 m)(220 rad)=9.90 m.𝑥=rθ=0.0450 m220 rad=9.90 m.

10.27

Discussion

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.

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/s2–300rad/s2. How long does it take the reel to come to a stop?

Strategy

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.

Solution

The equation states

ω=ω0+αt.𝜔=𝜔0+αt.

10.28

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

t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s.𝑡=𝜔−𝜔0𝛼=0−220 rad/s−300rad/s2=0.733 s.

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Discussion

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?

Strategy

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=xr.𝜃=𝑥𝑟.

10.30

Solving this equation for x𝑥 yields

x=rθ.𝑥=rθ.

10.31

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=(200rev)2πrad1 rev=1257rad.𝜃=200rev2πrad1 rev=1257rad.

10.32

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

x=rθ=(0.350 m)(1257 rad)=440m.𝑥=rθ=0.350 m1257 rad=440m.

10.33

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

10.34

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

ω==[0+2(0.250 rad/s2)(1257 rad)]1/225.1 rad/s.𝜔=0+2(0.250 rad/s2)(1257 rad)1/2=25.1 rad/s.

10.35

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

v=rω=(0.350 m)(25.1 rad/s)=8.77 m/s.𝑣=rω=0.350 m25.1 rad/s=8.77 m/s.

10.36

Discussion

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.

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.)

Strategy

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.

Solution

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

θ=ω−−t=(6.0 rpm)(2.0 min)=12 rev.𝜃=𝜔-𝑡=6.0 rpm2.0 min=12 rev.

10.37

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 rev)(2πrad1 rev)=75.4 rad.𝜃=12 rev2𝜋rad1 rev=75.4 rad.

10.38

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

x=rθ=(0.15 m)(75.4 rad)=11 m.𝑥=rθ=0.15 m75.4 rad=11 m.