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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. 

h3. Part A
 
What is the bucket's acceleration as it falls?

Solution:  We will consider two different methods to obtain the solution.

h4. 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}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.{note}

For the bucket, we write Newton's 2nd Law:

{latex}\begin{large} \[ mg - T = ma \] \end{large}{latex}

For the pulley, we sum the torques about the fixed axis defined by the axle:

{latex}\begin{large} \[ TR = I\alpha \] \end{large}{latex}

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:

{latex}\begin{large}\[ \alpha R = a \]\end{large}{latex}

With this assumption, we can solve the system of equations to obtain:

{latex}\begin{large} \[ a = \frac{g}{1+\frac{\displaystyle I}{\displaystyle mR^{2}}} \] \end{large}{latex}


h4. 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:

{latex}\begin{large}\[ L_{\rm axis} = m\vec{r}_{\rm cm,axis}\times\vec{v}_{\rm cm} \] \end{large}{latex}

The total angular momentum is:

{latex}\begin{large}\[ L_{\rm total} = I\omega + mvR \]\end{large}{latex}

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:

{latex}\begin{large}\[ \omega R = v \]\end{large}{latex}

We then set the sum of external torques equal to the change in angular momentum of the system:
{warning}Although Newton's 3rd Law guarantees that internal forces do not contribute to the change in momentum of the system's CM, the same is *not strictly true* for the change in angular momentum.  For an example of a case where ignoring the internal torques yields incorrect results, see [Atwood Revisited].{warning}

{note}The internal torques certainly cancel _in this case_ because the internal forces share the same [line of action], which guarantees that they have the same [lever arm] regardless of the axis chosen.  Together with Newton's 3rd Law (guaranteeing equal magnitudes and opposite directions for the internal forces) this implies equal and opposite internal torques.{note}

{latex}\begin{large}\[ mgR = \frac{d}{dt}\left(I \frac{v}{R} + mvR\right)\] \end{large}{latex}

Now, using the fact that 

{latex}\begin{large} \[ a = \frac{dv}{dt} \] \end{large}{latex}

lets us solve to find:

{latex}\begin{large} \[ a = \frac{g}{1 + \frac{\displaystyle I}{\displaystyle mR^{2}}} \] \end{large}{latex}

{tip}The same answer as was obtained via method 1.{tip}


h3. Part B

What is the bucket's speed after falling 5.0 m down the well.