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398 CHAPTER 13 Dynamics of a Rigid Body

A meterstick is initially standing vertically on the floor. If the
EXAMPLE 2
meterstick falls over, with what angular velocity will it hit the
y floor? Assume that the end in contact with the floor does not slip.
SOLUTION: The motion of the meterstick is rotation about a fixed axis passing
f
through the point of contact with the floor (see Fig. 13.3).The stick is a uniform
Initially, rod of mass M and length l 1.0 m. Its moment of inertia about the end is Ml 3
2
y = l /2. 1 2 2 2
1
(see Table 12.3), and its rotational kinetic energy is therefore I Ml >6.
2
When stick hits
y floor, y = 0. The gravitational potential energy is Mgy, where y is the height of the center of
2
mass above the floor. When the meterstick is standing vertically, the initial angu-
x lar velocity is 0 and y l 2, so the total energy is
O 1 1
2 2
1
FIGURE 13.3 Meterstick rotating about E Ml Mgy 0 Mgl>2 (13.11)
6
1
1
its lower end.
Just before the meterstick hits the floor, the angular velocity is and y 0.
2
2
The energy is
1
2 2
2 2
1
E Ml Mgy Ml 0 (13.12)
2
6
2
2
6
Conservation of energy therefore implies
1 2 2
2
6 Ml Mgl>2
from which we obtain
3g
2
(13.13)
2
l
Taking the square root of both sides, we find
3g 3 9.81m/s 2
5.4 radians/s
2 B
B l 1.0 m
At what instantaneous rate is gravity delivering energy to the
EXAMPLE 3
meterstick of Example 2 just before it hits the floor? The mass
of the meterstick is 0.15 kg.
SOLUTION: The rate of energy delivery is the power,
P t
From Example 2, we know 5.4 radians/s just before the stick hits the floor.
At that instant, gravity acts perpendicular to the stick at the center of mass
(in the next chapter we will see that the weight acts as if concentrated at the center
of mass), a distance R l 2 0.50 m from the end. So the torque exerted by
gravity is

l 2
t FR sin u mg sin 90 0.15 kg 9.81 m/s 0.50 m 1
2
0.74 N m

Thus the instantaneous power delivered by the torque due to gravity is
P t 0.74 N m 5.4 radians/s 4.0 W