Download PWE 8-12: A Simple Pulley I

Example 8-12 A Simple Pulley I
A pulley is a solid uniform cylinder of mass Mpulley and radius R that is free to rotate around an axis through its center. A lightweight rope is wound around the pulley. You exert a constant force of magnitude F on the rope, which makes the rope
unwind and rotates the pulley (Figure 8-25). The rope does not stretch and does
not slip on the pulley. What is the angular acceleration of the pulley about its axle?
Pulley
Lightweight rope
Figure 8-25 ​Exerting a torque on a pulley What is the angular acceleration of the pulley as the
F
force F makes the rope unwind?
Set Up
We begin by drawing a free-body diagram for
the pulley, taking care to draw each force at the
point where it acts. We are told that the rope is
lightweight (that is, it has much less mass than
the pulley), so the force that you exert on the
free end of the rope has the same magnitude F as
the force that the rope exerts on the pulley. This
force exerts a torque on the pulley and causes an
angular acceleration. We’ll use the rotational form
of Newton’s second law to determine this angular
acceleration.
Solve
The free-body diagram shows that only the
tension force F exerts a torque on the pulley
causing it to rotate around its axle. (The support
force Fsupport and the weight of the pulley Wpulley
both act at its rotation axis, so the lever arm is
zero for both of these forces.) The lever arm for
the tension force is R, so the tension torque is R
multiplied by F.
From Table 8-1 in Section 8-3, the ­moment of
inertia of a solid cylinder around its central axis
is I = MR2 >2. Insert this into Newton’s second
law for rotational motion and solve for apulley, z.
Reflect
Newton’s second law for rotational motion:
a tz = Ipulley apulley, z
(8-20)
Fsupport
wpulley
Torque due to tension force:
tz = r›F = RF
F
lever arm
R
(This torque makes the pulley rotate in the
clockwise direction, so we take clockwise to
be the positive rotation direction.)
This is the only torque acting on the pulley, so
a tz = RF
a tz = Ipulley apulley,z
rotation axis
F
Substitute values of a tz and Ipulley = MpulleyR2 >2:
1
RF = Mpulley R2apulley,z
2
2RF
2F
Solve for angular acceleration az: apulley,z =
=
Mpulley R
Mpulley R2
Our result for the angular acceleration apulley,z depends on the pulling force F, the pulley mass Mpulley, and the pulley
radius R. It makes sense that our result is proportional to the ratio F>Mpulley: A greater force F means a stronger pull and
a greater angular acceleration, while a greater mass Mpulley means the pulley is more difficult to rotate and gives a smaller
angular acceleration.
Our result also shows that the larger the pulley radius R, the smaller the angular acceleration apulley,z. This may seem
backwards, since a larger radius means that the force F causes a larger torque t = RF. However, increasing the ­radius
increases the moment of inertia I = Mpulley R2 >2 by a larger factor than it increases the torque. (Doubling the radius
doubles the torque but quadruples the moment of inertia.) So the moment of inertia plays a more important role, which
is why apulley,z decreases with increasing pulley radius.