9781118230725.pdf

280 CHAPTER 10 ROTATION

Additional examples, video, and practice available at WileyPLUS

KEY IDEA

Because the moment arm for Fis no longer zero, the torque

:

g

Checkpoint 7

The figure shows an overhead view of a meter stick that can pivot about the point indicated, which is

to the left of the stick’s midpoint. Two horizontal forces, and , are applied to the stick. Only is

shown. Force is perpendicular to the stick and is applied at the right end. If the stick is not to turn,

(a) what should be the direction of , and (b) should F F 2 be greater than, less than, or equal to F 1?

:

2

F

:

2

F

:

F 1

:

F 2

:

1

F 1

Pivot point

Sample Problem 10.09 Using Newton’s second law for rotation in a basic judo hip throw

To throw an 80 kg opponent with a basic judo hip throw, you
intend to pull his uniform with a force and a moment arm
d 1 0.30 m from a pivot point (rotation axis) on your right
hip (Fig. 10-18). You wish to rotate him about the pivot
point with an angular acceleration aof6.0 rad/s^2 —that is,
with an angular acceleration that is clockwisein the figure.
Assume that his rotational inertia Irelative to the pivot
point is 15 kg m^2.

(a) What must the magnitude of be if, before you throw
him, you bend your opponent forward to bring his center of
mass to your hip (Fig. 10-18a)?

KEY IDEA

We can relate your pull on your opponent to the given an-
gular acceleration avia Newton’s second law for rotation
(tnetIa).

Calculations:As his feet leave the floor, we can assume that
only three forces act on him: your pull , a force on him
from you at the pivot point (this force is not indicated in Fig.
10-18), and the gravitational force. To use tnetIa, we need
the corresponding three torques, each about the pivot point.
From Eq. 10-41 (t F), the torque due to your pull F

:

r

F

:

g

N

:

F

:

F

:

F

:

F

:

Figure 10-18A judo hip throw (a) correctly executed and (b) incor-

rectly executed.

Opponent's

center of

mass

Moment arm d 1

of your pull

Pivot

on hip

Moment arm d 2

of gravitational

force on

opponent

Moment

armd 1

of your pull

Fg Fg

(a) (b)

F

F

is equal to  F, where is the moment arm and the
sign indicates the clockwise rotation this torque tends to
cause. The torque due to is zero, because Nacts at the
:
N
:

d 1 d 1 r

pivot point and thus has moment arm 0.
To evaluate the torque due to , we can assume that
acts at your opponent’s center of mass. With the center of
mass at the pivot point,F has moment arm r0 and thus

:

g

F

:

F g

:

g

r

ponent is due to your pull , and we can write tnetIaas

d 1 FIa.
We then find

300 N. (Answer)

(b) What must the magnitude of be if your opponent
remains upright before you throw him, so that has a mo-
ment arm d 2 0.12 m (Fig. 10-18b)?

F

:

g

F

:

F

Ia

d 1



(15 kg m^2 )(6.0 rad/s^2 )

0.30 m

F

the torque due to F is zero. So, the only torque on your op-:

:

g

due to is now equal to d 2 mgand is positive because the

torque attempts counterclockwise rotation.

Calculations:Now we write tnetIaas

d 1 Fd 2 mgIa,

which gives

From (a), we know that the first term on the right is equal to

300 N. Substituting this and the given data, we have

613.6 N 610 N. (Answer)

The results indicate that you will have to pull much harder if

you do not initially bend your opponent to bring his center

of mass to your hip. A good judo fighter knows this lesson

from physics. Indeed, physics is the basis of most of the mar-

tial arts, figured out after countless hours of trial and error

over the centuries.



F300 N

(0.12 m)(80 kg)(9.8 m/s^2 )

0.30 m

F

Ia

d 1



d 2 mg

d 1

.

F

:

g