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 (from rest) 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.
Interactions: 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: [Rotation and Translation of a Rigid Body] and Point Particle Dynamics
Approach: We begin with free body diagrams for the two objects.
Note that the two forces acting on the bucket each have zero moment 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
[ m_
g - T = m_
a_
] \end
For the pulley, there will be no translation, only rotation about the center of mass. 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{\displaystyle I_{p}}{\displaystyle m_
R^
}} ] \end
Note that Ip/(mbR2) is not equal to 1/2. The ratio of the pulley's moment of intertia to the pulley's mass times radius squared is 1/2, but we have the ratio of the pulley's moment of inertia to the bucket's mass times the pulley's radius squared.
or, using the formula for the moment of inertia of a cylinder:
\begin
[ a_
= \frac
{1+\frac{\displaystyle m_{p}}{\displaystyle 2m_
}} = \mbox
^
]\end
As expected, the acceleration approaches g if the mass of the bucket goes to infinity, and it approaches zero if the mass of the pulley goes to infinity.
Method 2
System: An alternate approach is to combine the bucket and the pulley into a single system.
Interactions: 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: [1-D Angular Momentum and Torque]
Approach: The angular momentum of the system can be expressed by summing the angular momentum of the parts. The pulley's contribution is Ip ωp. 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_
+ m_
v_
R ]\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:
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 moment 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.
\begin
[ m_
gR = \frac
\left(I_
\frac{v_{b}}
+ m_
v_
R\right)] \end
Now, using the fact that
\begin
[ a_
= \frac{dv_{b}}
] \end
lets us solve to find:
\begin
[ a_
= \frac
{1 + \frac{\displaystyle I_{p}}{\displaystyle m_
R^
}} ] \end
The same answer as was obtained via method 1.
Part B
What is the bucket's speed after falling 5.0 m down the well?
Solution
Again, we will use two methods.
Method 1
System: Bucket as point particle.
Interactins: External influences from the earth (gravity) and the rope (tension).
Model: One-Dimensional Motion with Constant Acceleration.
Approach: To use this method, you must first find the acceleration of the bucket using one of the methods of Part A. Once that acceleration is in hand, the problem is reduced to kinematics. The most direct solution is obtained by using:
\begin
[v_
^
= v_{y,{\rm i}}^
+ 2 a_
(y-y_
) ] \end
If we assume y = 0 m at the height of release, then:
\begin
[ v_
= \pm \sqrt{2 a_
y} ] \end
The speed of the bucket, then, must be:
\begin
[ v_
= \sqrt{2 (\mbox
^
) (\mbox
)} = \mbox
] \end
Method 2
System: Treat the bucket, pulley and the earth as a single system.
Interactions: Internal interactions of gravity (conservative) and tension from the rope (non-conservative) plus an external influence from the normal force on the pulley (non-conservative).
Model: [Mechanical Energy and Non-Conservative Work].
Approach: There will be no net non-conservative work on this system, even though there are non-conservative forces present. The normal force is a non-conservative external force, but does no work since the pulley experiences no displacement and the normal force has no moment arm and so exerts no torque. The tension forces on the pulley and the block are non-conservative internal forces. The tension forces do perform work on the system, but the work done by the downward tension on the pulley and the upward tension on the block are equal in size and opposite in sign. Therefore, their contributions cancel each other to give zero net non-conservative work.
To see that the net work from tension is zero requires us to use two methods of calculating work. Using force and displacement, you can show that the work done on the bucket is
[
_
= - |Ty|]
where |y| is the distance fallen. To find the work done on the pulley, we must use
[ W_
= \tau (\theta - \theta_
)]
but since
[ \tau = |T|R ]
and (assuming the rope does not slip or stretch)
[ \theta - \theta_
= |y|/R ]
we find that
[ W_
= + |Ty| ]
.
We can now use the conservation of energy, which consists of three parts:
\begin
[ K_
+ K_
+ U_
= K_
+ K_
+ U_
] \end
If we set h= 0 to be at the initial position of the bucket, then we can write:
\begin
[ m_
g(-y_
) = \frac
m_
v_
^
+ \frac
I_
\omega_
^
+ m_
g(-y_
) + m_
g(-y_
) ] \end
Because we chose a coordinate system with the positive y-axis pointing down, we must be sure to include the negative sign on all y-positions when finding heights (which must necessarily increase when moving upward).
where ya denotes the (constant) vertical position of the center of mass of the pulley. Substituting using the relationship:
\begin
[ \omega_
R = v_
]\end
gives
\begin
[ m_
g y_
= \frac
m_
v_
^
+ \frac
I_
\left(\frac{v_{b}}
\right)^
]\end
We can now solve for vb (selecting the positive root since we are finding a speed):
\begin
[ v_
= \sqrt{\frac{2 g y_{b}}{1 + \frac{\displaystyle I_{p}}{\displaystyle m_
R^{2}}}} = \mbox
] \end
The same answer as was found using method 1.