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
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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