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404 CHAPTER 13 Dynamics of a Rigid Body

To find the force P, we need to examine the equation for the translational
motion. The net horizontal force is F P f. Hence the equation for the
net
translational motion of the wheel is
Ma P f

from which
2

P Ma f 25 kg 4.0 m/s 50 N
150 N

Thus, the force required to accelerate a rolling wheel is larger than the force required
for a wheel that slips on a frictionless surface without rolling—for such a wheel
2
the force would be only Ma 25 kg 4.0 m/s 100 N. Here, the additional
1 2 1
rotational inertia MR adds an additional amount f Ma to the required
2
2
3
force, so the total required force is that for sliding
2
without rolling.

A solid cylinder of mass M and radius R rolls down a sloping
EXAMPLE 7
(a)
ramp that makes an angle with the ground (see Fig. 13.9a).
What is the acceleration of the cylinder? Assume that the cylinder is uniform and
R rolls without slipping.
SOLUTION: Figure 13.9b shows the “free-body” diagram for the cylinder. The
forces on the cylinder are the normal force N exerted by the ramp, the friction
force f exerted by the ramp, and the weight w.The friction force is exerted on the
b rim of the cylinder, and the weight is effectively exerted at the center of the cylin-
der (in the next chapter we will see that the weight can always be regarded as con-
(b)
centrated at the center of mass). As axis of rotation, we take the axis that passes
y
through the center of the cylinder. The weight exerts no torque about this axis,
and neither does the normal force (zero moment arm). Hence, the only force that
N
Normal force acts exerts a torque is the friction force, and so
along a radial line,
exerting zero torque.
O
t Rf
f
b
x The equation of rotational motion is then
Only friction force Weight acts at
f exerts a torque w center of mass and
about axis of exerts no torque I R f
cylinder. about the axis.
The moment of inertia of a uniform cylinder is the same as that of a disk,
1
FIGURE 13.9 (a) A cylinder rolling down I MR 2 . Furthermore, for rolling motion without slipping, a R. Hence
2
an inclined plane. (b) “Free-body” diagram
for the cylinder. 1
2 MR a R f
or
2f
a (13.25)
M

To evaluate the acceleration, we need to eliminate the friction force f from this
equation. We can do this by appealing to the equation for the component of the
translational motion along the ramp (the motion along the x direction in Fig. 13.9b).