# Learning Objectives

### Learning Objectives

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

• 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 $θθ size 12{θ} {}$, $ωω size 12{ω} {}$, and $αα size 12{α} {}$ 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 $αα size 12{α} {}$ is large for a long period of time $tt size 12{α} {}$, then the final angular velocity $ωω size 12{ω} {}$ and angle of rotation $θθ size 12{θ} {}$ 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 $ωω size 12{ω} {}$, $αα size 12{α} {}$, and $tt size 12{t} {}$. To determine this equation, we recall the following kinematic equation for translational, or straight-line, motion.

10.17

Note that in rotational motion $a=ata=at size 12{a=a rSub { size 8{t} } } {}$, and we shall use the symbol $aa size 12{a} {}$ for tangential or linear acceleration from now on. As in linear kinematics, we assume $aa size 12{a} {}$ is constant, which means that angular acceleration $αα size 12{α} {}$ is also a constant, because $a=rαa=rα size 12{a=rα} {}$. Now, let us substitute $v=rωv=rω size 12{v=rω} {}$ and $a=rαa=rα size 12{a=rα} {}$ into the linear equation above using

10.18 $rω=rω0+rαt.rω=rω0+rαt. size 12{rω=rω rSub { size 8{0} } +rαt} {}$

The radius $rr size 12{r} {}$ cancels in the equation, yielding

10.19

where $ω0ω0 size 12{ω rSub { size 8{0} } } {}$ is the initial angular velocity. This last equation is a kinematic relationship among $ωω size 12{ω} {}$, $αα size 12{α} {}$, and $tt size 12{t} {}$ —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
$θ=ω¯tθ=ω¯t size 12{θ= {overline {ωt}} } {}$ $x=v-tx=v-t size 12{x= { bar {v}}t} {}$
$ω=ω0+αtω=ω0+αt size 12{ω=ω rSub { size 8{0} } +αt} {}$ $v=v0+atv=v0+at size 12{v=v rSub { size 8{0} } + ital "at"} {}$ (constant $αα size 12{α} {}$, $aa size 12{a} {}$)
$θ=ω0t+12αt2θ=ω0t+12αt2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {}$ $x=v0t+12at2x=v0t+12at2 size 12{x=v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}$ (constant $αα size 12{α} {}$, $aa size 12{a} {}$)
$ω2=ω02+2αθω2=ω02+2αθ size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} rSup { size 8{2} } } +2 ital "αθ"} {}$ $v2=v02+2axv2=v02+2ax$ (constant $αα$, $aa$)
Table 10.2 Rotational Kinematic Equations

In these equations, the subscript 0 denotes initial values ($θ0θ0 size 12{θ rSub { size 8{0} } } {}$, $x0x0 size 12{x rSub { size 8{0} } } {}$, and $t0t0 size 12{t rSub { size 8{0} } } {}$ are initial values), and the average angular velocity $ω-ω- size 12{ { bar {ω}}} {}$ and average velocity $v-v- size 12{ { bar {v}}} {}$ are defined as follows

10.20

The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which $aa size 12{a} {}$ and $αα size 12{α} {}$ are constant.

### Problem-solving Strategy for Rotational Kinematics

1. Examine the situation to determine that rotational kinematics or 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.

### Example 10.3Calculating 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 $110rad/s2110rad/s2 size 12{"110""rad/s" rSup { size 8{2} } } {}$ 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 s elapse?

(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 $αα size 12{α} {}$ and $tt size 12{α} {}$ are given and $ωω size 12{ω} {}$ needs to be determined. The most straightforward equation to use is $ω=ω0+αtω=ω0+αt size 12{ω=ω rSub { size 8{0} } +αt} {}$ because the unknown is already on one side and all other terms are known. That equation states that

10.21 $ω=ω0+αt.ω=ω0+αt. size 12{ω=ω rSub { size 8{0} } +αt"."} {}$

We are also given that $ω0=0ω0=0 size 12{ω rSub { size 8{0} } =0} {}$ (it starts from rest), so that

10.22 $ω=0+110 rad/s22 s=220rad/s.ω=0+110 rad/s22 s=220rad/s. size 12{ω=0+ left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right )="220 rad/s."} {}$

Solution for (b)

Now that $ωω size 12{ω} {}$ is known, the speed $vv size 12{v} {}$ can most easily be found using the relationship

10.23 $v=rω,v=rω, size 12{v=rω","} {}$

where the radius $rr size 12{α} {}$ of the reel is given to be 4.50 cm; thus,

10.24 $v=0.0450 m220 rad/s=9.90 m/s.v=0.0450 m220 rad/s=9.90 m/s. size 12{v= left (0 "." "0450"" m" right ) left ("220"" rad/s" right )=9 "." "90"" m/s."} {}$

Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have $m*rad=mm*rad=m size 12{m times "rad"=m} {}$.

Solution for (c)

Here, we are asked to find the number of revolutions. Because $1 rev=2π rad1 rev=2π rad size 12{1" rev"=2π" rad"} {}$, we can find the number of revolutions by finding $θθ size 12{θ} {}$ in radians. We are given $αα size 12{α} {}$ and $tt size 12{t} {}$, and we know $ω0ω0 size 12{ω rSub { size 8{ {} rSub { size 6{0} } } } } {}$ is zero, so that $θθ size 12{θ} {}$ can be obtained using $θ=ω0t+12αt2θ=ω0t+12αt2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {}$.

10.25 θ=ω0t+12αt2=0+0.500110 rad/s22 s2=220 radθ=ω0t+12αt2=0+0.500110 rad/s22 s2=220 radalignl { stack { size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {} # " "=0+ left (0 "." "500" right ) left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right ) rSup { size 8{2} } ="220"" rad" {} } } {}

10.26 $θ=220 rad1 rev2π rad=35.0 rev.θ=220 rad1 rev2π rad=35.0 rev. size 12{θ= left ("220"" rad" right ) { {1" rev"} over {2π" rad"} } ="35" "." 0" rev."} {}$

Solution for (d)

The number of meters of fishing line is $xx size 12{x} {}$, which can be obtained through its relationship with $θθ size 12{θ} {}$

10.27 $x=rθ=0.0450 m220 rad=9.90 m.x=rθ=0.0450 m220 rad=9.90 m. size 12{x=rθ= left (0 "." "0450"" m" right ) left ("220"" rad" right )=9 "." "90"" m"} {}$

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 2,100 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 10.8 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 10.4Calculating 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 $–300rad/s2–300rad/s2 size 12{"300""rad/s" rSup { size 8{2} } } {}$. How long does it take the reel to come to a stop?

Strategy

We are asked to find the time $tt size 12{α} {}$ 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 $ω0=220 rad/sω0=220 rad/s size 12{ω rSub { size 8{0} } ="220"" rad/s"} {}$ and the final angular velocity $ωω size 12{ω} {}$ is zero. The angular acceleration is given to be $α=−300rad/s2α=−300rad/s2 size 12{α= - "300" "rad/s" rSup { size 8{2} } } {}$. Examining the available equations, we see all quantities but t are known in $ω=ω0+αt,ω=ω0+αt, size 12{ω=ω rSub { size 8{0} } +αt} {}$ making it easiest to use this equation.

Solution

The equation states

10.28 $ω=ω0+αt.ω=ω0+αt. size 12{ω=ω rSub { size 8{0} } +αt"."} {}$

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

10.29 $t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s.t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s. size 12{t= { {ω - ω rSub { size 8{0} } } over {α} } = { {0 - "220"" rad/s"} over { - "300""rad/s" rSup { size 8{2} } } } =0 "." "733"" s."} {}$

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 10.5Calculating 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 $0.250rad/s20.250rad/s2 size 12{0 "." "250""rad/s" rSup { size 8{2} } } {}$. After the wheels have made 200 revolutions with 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 $xx size 12{x} {}$, and in (b) we are asked to find $ωω size 12{ω} {}$ and $vv size 12{v} {}$. We are given the number of revolutions $θθ size 12{θ} {}$, the radius of the wheels $rr size 12{r} {}$, and the angular acceleration $αα size 12{α} {}$.

Solution for (a)

The distance $xx size 12{x} {}$ is very easily found from the relationship between distance and rotation angle.

10.30 $θ=xrθ=xr size 12{θ= { {x} over {r} } } {}$

Solving this equation for $xx size 12{x} {}$ yields

10.31 $x=rθ.x=rθ. size 12{x=rθ.} {}$

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.

10.32 $θ=200rev2πrad1 rev=1257radθ=200rev2πrad1 rev=1257rad size 12{θ= left ("200"" rev" right ) { {2π" rad"} over {"1 rev"} } ="1257"" rad"} {}$

Now we can substitute the known values into $x=rθx=rθ size 12{x=rθ} {}$ to find the distance the train moved down the track.

10.33 $x=rθ=0.350 m1257 rad=440 mx=rθ=0.350 m1257 rad=440 m size 12{x=rθ= left (0 "." "350"`m right ) left ("1257"" rad" right )="440"" m"} {}$

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 $ωω$.

10.34 $ω2=ω02+2αθω2=ω02+2αθ$

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

10.35

We can find the linear velocity of the train, $vv size 12{v} {}$, through its relationship to $ωω size 12{ω} {}$.

10.36 $v=rω=0.350 m25.1 rad/s=8.77 m/sv=rω=0.350 m25.1 rad/s=8.77 m/s size 12{v=rω= left (0 "." "350"" m" right ) left ("25" "." 1" rad/s" right )=8 "." "77"" m/s"} {}$

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 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels.

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

### Example 10.6Calculating 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 rpm, calculate the total distance traveled by the fly during a 2 min cooking period. Ignore the start-up and slow-down times.

Strategy

First, find the total number of revolutions $θθ size 12{θ} {}$, and then the linear distance $xx size 12{x} {}$ traveled. $θ=ω¯tθ=ω¯t size 12{θ= {overline {ωt}} } {}$ can be used to find $θθ size 12{θ} {}$ because $ω-ω- size 12{ { bar {ω}}} {}$ is given to be 6.0 rpm.

Solution

Entering known values into $θ=ω¯tθ=ω¯t size 12{θ= {overline {ωt}} } {}$ gives

10.37 $θ=ω-t=6.0 rpm2.0 min=12 rev.θ=ω-t=6.0 rpm2.0 min=12 rev.$

As always, it is necessary to convert revolutions to radians before calculating a linear quantity like $xx size 12{x} {}$ from an angular quantity like $θθ size 12{θ} {}$.

10.38 $θ=12 rev2π rad1 rev=75.4 radθ=12 rev2π rad1 rev=75.4 rad size 12{θ= left ("12"" rev" right ) left ( { {2π" rad"} over {"1 rev"} } right )="75" "." 4" rad"} {}$

Now, using the relationship between $xx size 12{x} {}$ and $θθ size 12{θ} {}$, we can determine the distance traveled.

10.39

Discussion

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.