9781118230725.pdf

284 CHAPTER 10 ROTATION

Sample Problem 10.11 Work, rotational kinetic energy, torque, disk

Let the disk in Fig. 10-19 start from rest at time t0 and

also let the tension in the massless cord be 6.0 N and the an-

gular acceleration of the disk be 24 rad/s^2. What is its rota-

tional kinetic energy Katt2.5 s?

KEY IDEA

We can find Kwith Eq. 10-34 (K^12 Iv^2 ).We already know

Calculations: First, we relate the change in the kinetic

energy of the disk to the net work Wdone on the disk, using

the work – kinetic energy theorem of Eq. 10-52 (KfKiW).

With Ksubstituted for Kfand 0 for Ki, we get

KKiW 0 WW. (10-60)

Next we want to find the work W. We can relate Wto

the torques acting on the disk with Eq. 10-53 or 10-54. The

only torque causing angular acceleration and doing work is

the torque due to force on the disk from the cord, which isT

:

that , but we do not yet know vat t2.5 s.
However, because the angular acceleration ahas the con-
stant value of 24 rad/s^2 , we can apply the equations for
constant angular acceleration in Table 10-1.

Calculations: Because we want vand know aandv 0 (0),
we use Eq. 10-12:

vv 0 at 0 atat.

Substitutingvatand into Eq. 10-34, we find

(Answer)

KEY IDEA

We can also get this answer by finding the disk’s kinetic
energy from the work done on the disk.

90 J.

^14 (2.5 kg)[(0.20 m)(24 rad/s^2 )(2.5 s)]^2

K^12 Iv^2 ^12 (^12 MR^2 )(at)^2 ^14 M(Rat)^2

I^12 MR^2

I^12 MR^2

Additional examples, video, and practice available at WileyPLUS

circular path, only the tangential component Ftof the force accelerates the parti-

cle along the path. Therefore, only Ftdoes work on the particle. We write that

workdWasFtds. However, we can replace dswithr du, where duis the angle

through which the particle moves. Thus we have

dWFtr du. (10-58)

From Eq. 10-40, we see that the product Ftris equal to the torque t, so we can

rewrite Eq. 10-58 as

dWtdu. (10-59)

The work done during a finite angular displacement from uitoufis then

which is Eq. 10-53. It holds for any rigid body rotating about a fixed axis.

Equation 10-54 comes directly from Eq. 10-53.

We can find the power Pfor rotational motion from Eq. 10-59:

which is Eq. 10-55.

P

dW

dt

t

du

dt

tv,

W

uf

ui

tdu,

equal to TR. Because ais constant, this torque also must

be constant. Thus, we can use Eq. 10-54 to write

Wt(ufui)TR(ufui). (10-61)

Because a is constant, we can use Eq. 10-13 to find

ufui. With vi0, we have

.

Now we substitute this into Eq. 10-61 and then substitute the

result into Eq. 10-60. Inserting the given values T6.0 N

anda24 rad/s^2 , we have

90 J. (Answer)

^12 (6.0 N)(0.20 m)(24 rad/s^2 )(2.5 s)^2

KWTR(ufui)TR(^12 at^2 )^12 TRat^2

ufuivit^12 at^2  0 ^12 at^2 ^12 at^2