284
CHAPTE R 10 ROTATION
Sample Problem 10.11 Work, rotational kinetic energy, torque, disk
Let the disk in Fig. 10-19 start from rest at time t ϭ 0 and
also let the tension in the massless cord be 6.0 N and the angular acceleration of the disk be Ϫ24 rad/s
2
. What is its rotational kinetic energy K at t ϭ 2.5 s?
KEY IDEA
We can find K with Eq. 10-34
We already know
(K ϭ
1
2 Iv
2
).
Calculations: First, we relate the change in the kinetic
energy of the disk to the net work W done on the disk, using
the work – kinetic energy theorem of Eq. 10-52 (K f Ϫ K i ϭ W).
With K substituted for K f and 0 for K i , we get
K ϭ K i ϩ W ϭ 0 ϩ W ϭ W.
( 1 0 - 6 0 )
Next we want to find the work W. We can relate W to
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 is
T
:
that
, but we do not yet know v at t ϭ 2.5 s.
However, because the angular acceleration a has the constant value of Ϫ24 rad/s
2
, we can apply the equations for
constant angular acceleration in Table 10-1.
Calculations: Because we want v and know a and v 0 (ϭ 0),
we use Eq. 10-12:
v ϭ v 0 ϩ at ϭ 0 ϩ at ϭ at.
Substituting v ϭ at and
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.
ϭ 90 J.
ϭ
1
4 (2.5 kg)[(0.20 m)(Ϫ24 rad/s
2
)(2.5 s)]
2
K ϭ
1
2 Iv
2 ϭ
1
2 (
1
2 MR
2
)(at)
2 ϭ
1
4 M(Rat)
2
I ϭ
1
2 MR
2
I ϭ
1
2 MR
2
Additional examples, video, and practice available at WileyPLUS
circular path, only the tangential component F t of the force accelerates the particle along the path. Therefore, only F t does work on the particle. We write that
work dW as F t ds. However, we can replace ds with r du, where du is the angle
through which the particle moves. Thus we have
dW ϭ F t r du.
(10-58)
From Eq. 10-40, we see that the product F t r is equal to the torque t, so we can
rewrite Eq. 10-58 as
dW ϭ t du.
(10-59)
The work done during a finite angular displacement from u i to u f is 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 P for rotational motion from Eq. 10-59:
which is Eq. 10-55.
P ϭ
dW
dt
ϭ t
du
dt
ϭ tv,
W ϭ ͵
u f
u i
t du,
equal to ϪTR. Because a is constant, this torque also must
be constant. Thus, we can use Eq. 10-54 to write
W ϭ t(u f Ϫ u i ) ϭ ϪTR(u f Ϫ u i ).
(10-61)
Because a is constant, we can use Eq. 10-13 to find
u f Ϫ u i . With v i ϭ 0, we have
.
Now we substitute this into Eq. 10-61 and then substitute the
result into Eq. 10-60. Inserting the given values T ϭ 6.0 N
and a ϭ Ϫ24 rad/s
2
, we have
(Answer)
ϭ 90 J.
ϭ Ϫ
1
2 (6.0 N)(0.20 m)(Ϫ24 rad/s
2
)(2.5 s)
2
K ϭ W ϭ ϪTR(u f Ϫ u i ) ϭ ϪTR(
1
2 at
2
) ϭ Ϫ
1
2 TRat
2
u f Ϫ u i ϭ v i t ϩ
1
2 at
2 ϭ 0 ϩ
1
2 at
2 ϭ
1
2 at
2
CHAPTE R 10 ROTATION
Sample Problem 10.11 Work, rotational kinetic energy, torque, disk
Let the disk in Fig. 10-19 start from rest at time t ϭ 0 and
also let the tension in the massless cord be 6.0 N and the angular acceleration of the disk be Ϫ24 rad/s
2
. What is its rotational kinetic energy K at t ϭ 2.5 s?
KEY IDEA
We can find K with Eq. 10-34
We already know
(K ϭ
1
2 Iv
2
).
Calculations: First, we relate the change in the kinetic
energy of the disk to the net work W done on the disk, using
the work – kinetic energy theorem of Eq. 10-52 (K f Ϫ K i ϭ W).
With K substituted for K f and 0 for K i , we get
K ϭ K i ϩ W ϭ 0 ϩ W ϭ W.
( 1 0 - 6 0 )
Next we want to find the work W. We can relate W to
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 is
T
:
that
, but we do not yet know v at t ϭ 2.5 s.
However, because the angular acceleration a has the constant value of Ϫ24 rad/s
2
, we can apply the equations for
constant angular acceleration in Table 10-1.
Calculations: Because we want v and know a and v 0 (ϭ 0),
we use Eq. 10-12:
v ϭ v 0 ϩ at ϭ 0 ϩ at ϭ at.
Substituting v ϭ at and
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.
ϭ 90 J.
ϭ
1
4 (2.5 kg)[(0.20 m)(Ϫ24 rad/s
2
)(2.5 s)]
2
K ϭ
1
2 Iv
2 ϭ
1
2 (
1
2 MR
2
)(at)
2 ϭ
1
4 M(Rat)
2
I ϭ
1
2 MR
2
I ϭ
1
2 MR
2
Additional examples, video, and practice available at WileyPLUS
circular path, only the tangential component F t of the force accelerates the particle along the path. Therefore, only F t does work on the particle. We write that
work dW as F t ds. However, we can replace ds with r du, where du is the angle
through which the particle moves. Thus we have
dW ϭ F t r du.
(10-58)
From Eq. 10-40, we see that the product F t r is equal to the torque t, so we can
rewrite Eq. 10-58 as
dW ϭ t du.
(10-59)
The work done during a finite angular displacement from u i to u f is 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 P for rotational motion from Eq. 10-59:
which is Eq. 10-55.
P ϭ
dW
dt
ϭ t
du
dt
ϭ tv,
W ϭ ͵
u f
u i
t du,
equal to ϪTR. Because a is constant, this torque also must
be constant. Thus, we can use Eq. 10-54 to write
W ϭ t(u f Ϫ u i ) ϭ ϪTR(u f Ϫ u i ).
(10-61)
Because a is constant, we can use Eq. 10-13 to find
u f Ϫ u i . With v i ϭ 0, we have
.
Now we substitute this into Eq. 10-61 and then substitute the
result into Eq. 10-60. Inserting the given values T ϭ 6.0 N
and a ϭ Ϫ24 rad/s
2
, we have
(Answer)
ϭ 90 J.
ϭ Ϫ
1
2 (6.0 N)(0.20 m)(Ϫ24 rad/s
2
)(2.5 s)
2
K ϭ W ϭ ϪTR(u f Ϫ u i ) ϭ ϪTR(
1
2 at
2
) ϭ Ϫ
1
2 TRat
2
u f Ϫ u i ϭ v i t ϩ
1
2 at
2 ϭ 0 ϩ
1
2 at
2 ϭ
1
2 at
2