SECTION 10.7 • Relationship Between Torque and Angular Acceleration
309
Quick Quiz 10.10
You turn off your electric drill and find that the time in-
terval for the rotating bit to come to rest due to frictional torque in the drill is $t. You
replace the bit with a larger one that results in a doubling of the moment of inertia of
the entire rotating mechanism of the drill. When this larger bit is rotated at the same
angular speed as the first and the drill is turned off, the frictional torque remains the
same as that for the previous situation. The time for this second bit to come to rest is
(a) 4$t (b) 2$t (c) $t (d) 0.5$t (e) 0.25$t (f) impossible to determine.
Example 10.10 Rotating Rod
A uniform rod of length L and mass M is attached at one
end to a frictionless pivot and is free to rotate about the
pivot in the vertical plane, as in Figure 10.18. The rod is re-
leased from rest in the horizontal position. What is the ini-
tial angular acceleration of the rod and the initial linear ac-
celeration of its right end?
Solution We cannot use our kinematic equations to find (
or a because the torque exerted on the rod varies with its
angular position and so neither acceleration is constant. We
have enough information to find the torque, however, which
we can then use in Equation 10.21 to find the initial ( and
then the initial a.
The only force contributing to the torque about an axis
through the pivot is the gravitational force M
g exerted on
the rod. (The force exerted by the pivot on the rod has zero
torque about the pivot because its moment arm is zero.) To
compute the torque on the rod, we assume that the gravita-
tional force acts at the center of mass of the rod, as shown in
Figure 10.18. The magnitude of the torque due to this force
about an axis through the pivot is
With 2 " I(, and
for this axis of rotation (see
Table 10.2), we obtain
All points on the rod have this initial angular acceleration.
To find the initial linear acceleration of the right end of
the rod, we use the relationship a
t
"
r( (Eq. 10.11), with
r " L:
3
2
g
a
t
"
L( "
3g
2L
(1)
( "
2
I
"
Mg(L/2)
1
3
ML
2
"
I "
1
3
ML
2
%
2 "
Mg
#
L
2
$
What If?
What if we were to place a penny on the end of
the rod and release the rod? Would the penny stay in contact
with the rod?
Answer The result for the initial acceleration of a point
on the end of the rod shows that a
t
-
g. A penny will fall at
acceleration g. This means that if we place a penny at the
end of the rod and then release the rod, the end of the rod
falls faster than the penny does! The penny does not stay
in contact with the rod. (Try this with a penny and a meter
stick!)
This raises the question as to the location on the rod at
which we can place a penny that will stay in contact as both
begin to fall. To find the linear acceleration of an arbitrary
point on the rod at a distance r * L from the pivot point, we
combine (1) with Equation 10.11:
For the penny to stay in contact with the rod, the limiting
case is that the linear acceleration must be equal to that due
to gravity:
Thus, a penny placed closer to the pivot than two thirds of
the length of the rod will stay in contact with the falling rod
while a penny farther out than this point will lose contact.
r "
2
3
L
a
t
"
g "
3g
2L
r
a
t
"
r
( "
3g
2L
r
L
Pivot
Mg
Figure 10.18 (Example 10.10) A rod is free to rotate around a
pivot at the left end.