A bucket for collecting water from a well is suspended by a rope which is wound around a pulley. The empty bucket has a mass of 2.0 kg, and the pulley is essentially a uniform cylinder of mass 3.0 kg on a frictionless axle. Suppose a person drops the bucket into the well.
Part A
What is the bucket's acceleration as it falls?
Solution: We will consider two different methods to obtain the solution.
Method 1
Systems: The pulley and the bucket are treated as separate objects. The bucket can be treated as a point particle, but the pulley must be treated as a rigid body. The pulley and the bucket are each subject to external influences from the rope and from gravity. The pulley is also subject to a normal force from the axle.
Model: [Fixed-Axis Rotation] and Point Particle Dynamics
Approach: We begin with free body diagrams for the two objects.
PICTURE
Note that the two forces acting on the bucket each have zero [lever arm] relative to the center of mass of the bucket. Thus, they have no tendency to produce rotation about the center of mass and so we can justifiably treat the bucket as a point particle.
For the bucket, we write Newton's 2nd Law:
\begin
[ mg - T = ma ] \end
For the pulley, we sum the torques about the fixed axis defined by the axle:
\begin
[ TR = I\alpha ] \end
We now make the assumption that the rope does not stretch or slip as it unwinds. These assumptions allow us to relate the rotation rate of the pulley to the motion of the bucket:
\begin
[ \alpha R = a ]\end
With this assumption, we can solve the system of equations to obtain:
\begin
[ a = \frac
{1+\frac
{mR^
}} ] \end
Method 2
System: An alternate approach is to combine the bucket and the pulley into a single system. The system is subject to external forces from the earth acting on the bucket and the pulley and from the normal force acting on the pulley.
Model: [Fixed-Axis Rotation]
Approach: The angular momentum of the system can be expressed by summing the angular momentum of the parts. The pulley's contribution is I ω. The bucket can effectively be modeled as a point particle, so it will contribute:
\begin
[ L_
= m\vec
_
\times\vec
_
] \end
The total angular momentum is:
\begin
[ L_
= I\omega + mvR ]\end
We now assume that the rope does not stretch or slip, allowing us to relate the rotational speed of the pulley to the speed of the bucket as it falls:
\begin
[ \omega R = v ]\end
We then set the sum of external torques equal to the change in angular momentum of the system:
\begin
[ mgR = \frac
\left(I \frac
+ mvR\right)] \end
Now, using the fact that
\begin
[ a = \frac
] \end
lets us solve to find:
\begin
[ a = \frac
{1 + \frac
{mR^
}} ] \end
The same answer as was obtained via method 1.