271
10-4 KINETIC ENERGY OF ROTATION
Additional examples, video, and practice available at WileyPLUS
The radial and tangential accelerations are perpendicular to each other and form the components of the rider’s
acceleration (Fig. 10-10b). The magnitude of is given by
a ϭ
(10-29)
39.9 m/s
2
,
( A n s w e r )
or 4.1g (which is really exciting!). All these values are
acceptable.
To find the orientation of , we can calculate the angle u
shown in Fig. 10-10b:
tan u ϭ
However, instead of substituting our numerical results, let’s
use the algebraic results from Eqs. 10-27 and 10-28:
u ϭ tan
Ϫ1
.
(10-30)
The big advantage of solving for the angle algebraically is that
we can then see that the angle (1) does not depend on the
ring’s radius and (2) decreases as t goes from 0 to 2.20 s. That
is, the acceleration vector swings toward being radially inward because the radial acceleration (which depends on t
4
)
quickly dominates over the tangential acceleration (which
depends on only t).At our given time t ϭ 2.20 s, we have
u
Њ. (Answer)
ϭ tan
Ϫ1
2
3(6.39 ϫ 10
Ϫ2
rad/s
3
)(2.20 s)
3
ϭ 44.4
a
:
6ctr
9c
2
t
4
r ϭ tan
Ϫ1
2
3ct
3
a t
a r
.
a
:
Ϸ
ϭ 2(28.49 m/s
2
)
2 ϩ (27.91 m/s
2
)
2
2a
2
r ϩ a
2
t
a
:
a
:
Although this is fast (111 km/h or 68.7 mi/h), such speeds are
common in amusement parks and not alarming because (as
mentioned in Chapter 2) your body reacts to accelerations but
not to velocities. (It is an accelerometer, not a speedometer.)
From Eq. 10-26 we see that the linear speed is increasing as the
square of the time (but this increase will cut off at t ϭ 2.30 s).
Next, let’s tackle the angular acceleration by taking the
time derivative of Eq. 10-25:
a ϭ
(3ct
2
) ϭ 6ct
ϭ 6(6.39 ϫ 10
Ϫ2
rad/s
3
)(2.20 s) ϭ 0.843 rad/s
2
. (Answer)
The tangential acceleration then follows from Eq. 10-22:
a t ϭ ar ϭ 6ctr
(10-27)
ϭ 6(6.39 ϫ 10
Ϫ2
rad/s
3
)(2.20 s)(33.1 m)
ϭ 27.91 m/s
2
27.9 m/s
2
,
( A n s w e r )
or 2.8g (which is reasonable and a bit exciting). Equation
10-27 tells us that the tangential acceleration is increasing
with time (but it will cut off at t ϭ 2.30 s). From Eq. 10-23,
we write the radial acceleration as
a r ϭ v
2
r.
Substituting from Eq. 10-25 leads us to
a r ϭ (3ct
2
)
2
r ϭ 9c
2
t
4
r
(10-28)
ϭ 9(6.39 ϫ 10
Ϫ2
rad/s
3
)
2
(2.20 s)
4
(33.1 m)
ϭ 28.49 m/s
2
28.5 m/s
2
,
(Answer)
or 2.9g (which is also reasonable and a bit exciting).
Ϸ
Ϸ
dv
dt
ϭ
d
dt
10-4 KINETIC ENERGY OF ROTATION
After reading this module, you should be able to . . .
10.17 Find the rotational inertia of a particle about a point.
10.18 Find the total rotational inertia of many particles moving
around the same fixed axis.
10.19 Calculate the rotational kinetic energy of a
body in terms of its rotational inertia and its angular
speed.
● The kinetic energy K of a rigid body rotating about a fixed
axis is given by
(radian measure),
K ϭ
1
2 Iv
2
in which I is the rotational inertia of the body, defined as
for a system of discrete particles.
I ϭ ͚ m i r i
2
Learning Objectives
Key Idea
Kinetic Energy of Rotation
The rapidly rotating blade of a table saw certainly has kinetic energy due to that
rotation. How can we express the energy? We cannot apply the familiar formula
to the saw as a whole because that would give us the kinetic energy
only of the saw’s center of mass, which is zero.
K ϭ
1
2 mv
2
10-4 KINETIC ENERGY OF ROTATION
Additional examples, video, and practice available at WileyPLUS
The radial and tangential accelerations are perpendicular to each other and form the components of the rider’s
acceleration (Fig. 10-10b). The magnitude of is given by
a ϭ
(10-29)
39.9 m/s
2
,
( A n s w e r )
or 4.1g (which is really exciting!). All these values are
acceptable.
To find the orientation of , we can calculate the angle u
shown in Fig. 10-10b:
tan u ϭ
However, instead of substituting our numerical results, let’s
use the algebraic results from Eqs. 10-27 and 10-28:
u ϭ tan
Ϫ1
.
(10-30)
The big advantage of solving for the angle algebraically is that
we can then see that the angle (1) does not depend on the
ring’s radius and (2) decreases as t goes from 0 to 2.20 s. That
is, the acceleration vector swings toward being radially inward because the radial acceleration (which depends on t
4
)
quickly dominates over the tangential acceleration (which
depends on only t).At our given time t ϭ 2.20 s, we have
u
Њ. (Answer)
ϭ tan
Ϫ1
2
3(6.39 ϫ 10
Ϫ2
rad/s
3
)(2.20 s)
3
ϭ 44.4
a
:
6ctr
9c
2
t
4
r ϭ tan
Ϫ1
2
3ct
3
a t
a r
.
a
:
Ϸ
ϭ 2(28.49 m/s
2
)
2 ϩ (27.91 m/s
2
)
2
2a
2
r ϩ a
2
t
a
:
a
:
Although this is fast (111 km/h or 68.7 mi/h), such speeds are
common in amusement parks and not alarming because (as
mentioned in Chapter 2) your body reacts to accelerations but
not to velocities. (It is an accelerometer, not a speedometer.)
From Eq. 10-26 we see that the linear speed is increasing as the
square of the time (but this increase will cut off at t ϭ 2.30 s).
Next, let’s tackle the angular acceleration by taking the
time derivative of Eq. 10-25:
a ϭ
(3ct
2
) ϭ 6ct
ϭ 6(6.39 ϫ 10
Ϫ2
rad/s
3
)(2.20 s) ϭ 0.843 rad/s
2
. (Answer)
The tangential acceleration then follows from Eq. 10-22:
a t ϭ ar ϭ 6ctr
(10-27)
ϭ 6(6.39 ϫ 10
Ϫ2
rad/s
3
)(2.20 s)(33.1 m)
ϭ 27.91 m/s
2
27.9 m/s
2
,
( A n s w e r )
or 2.8g (which is reasonable and a bit exciting). Equation
10-27 tells us that the tangential acceleration is increasing
with time (but it will cut off at t ϭ 2.30 s). From Eq. 10-23,
we write the radial acceleration as
a r ϭ v
2
r.
Substituting from Eq. 10-25 leads us to
a r ϭ (3ct
2
)
2
r ϭ 9c
2
t
4
r
(10-28)
ϭ 9(6.39 ϫ 10
Ϫ2
rad/s
3
)
2
(2.20 s)
4
(33.1 m)
ϭ 28.49 m/s
2
28.5 m/s
2
,
(Answer)
or 2.9g (which is also reasonable and a bit exciting).
Ϸ
Ϸ
dv
dt
ϭ
d
dt
10-4 KINETIC ENERGY OF ROTATION
After reading this module, you should be able to . . .
10.17 Find the rotational inertia of a particle about a point.
10.18 Find the total rotational inertia of many particles moving
around the same fixed axis.
10.19 Calculate the rotational kinetic energy of a
body in terms of its rotational inertia and its angular
speed.
● The kinetic energy K of a rigid body rotating about a fixed
axis is given by
(radian measure),
K ϭ
1
2 Iv
2
in which I is the rotational inertia of the body, defined as
for a system of discrete particles.
I ϭ ͚ m i r i
2
Learning Objectives
Key Idea
Kinetic Energy of Rotation
The rapidly rotating blade of a table saw certainly has kinetic energy due to that
rotation. How can we express the energy? We cannot apply the familiar formula
to the saw as a whole because that would give us the kinetic energy
only of the saw’s center of mass, which is zero.
K ϭ
1
2 mv
2