Finding Displacement Using Impulse Momentum: A Problem with Monkeys and a Pulley

In summary, the conversation was about a problem involving dynamics of rigid bodies, specifically using impulse momentum to find the displacement of a monkey holding onto a rope while another monkey travels down to get a peanut. The conversation also mentioned the use of energy conservation principles as an alternative method to solve the problem.
  • #1
sgtserious
2
0
I am having a really hard time with dynamics of rigid bodies, but i think doing this problem will clear up some confusion i have.

Homework Statement


A massless rope hanging over a frictionless pulley of mass M supports two monkeys (one of mass M, the other of mass 2M). The system is released at rest at t = 0. During the following 2 sec, monkey B travels down 15 ft of rope to obtain a massless peanut at P. Monkey A holds tightly to the rope during these two sec. Find the displacement of A during the time interval. Treat the pulley as a uniform cylinder of radius R.
Relevant picture: diagram

Homework Equations


As mentioned in the topic title, the problem has to do with impulse momentum:
∫ƩFy = m(y'f - y'i) [for monkey A]
∫ƩFy = m(y'f - y'i) [for monkey B]
∫ƩM = (I[itex]^{c}_{zz}[/itex]w)[itex]_{f}[/itex] -(I[itex]^{c}_{zz}[/itex]w)[itex]_{i}[/itex] [for the pulley of mass M]

The Attempt at a Solution


This is what I tried to do
(all initial velocities and angular velocities are zero)

For monkey A:
∫[itex]^{2}_{0}[/itex](T[itex]_{1}[/itex] - Mg) dt = M*y'[itex]_{A}[/itex][itex]_{f}[/itex]
-> T[itex]_{1}[/itex]*t - Mgt = M*y'[itex]_{A}[/itex][itex]_{f}[/itex]
-> 2T[itex]_{1}[/itex] -2Mg = M*y'[itex]_{A}[/itex][itex]_{f}[/itex]

For monkey B:
∫[itex]^{2}_{0}[/itex](T[itex]_{2}[/itex] - 2Mg) dt = 2M*y'[itex]_{B}[/itex][itex]_{f}[/itex]
-> T[itex]_{2}[/itex]*t - 2Mgt = 2M*y'[itex]_{B}[/itex][itex]_{f}[/itex]
-> 2T[itex]_{2}[/itex] - 4Mg = 2M*y'[itex]_{B}[/itex][itex]_{f}[/itex]

For the pulley:
∫[itex]^{2}_{0}[/itex](MgR - 2MgR) dt = (I[itex]^{c}_{zz}[/itex]w)[itex]_{f}[/itex]
-> -2MgR = (1/2)MR[itex]^{2}[/itex]w[itex]_{f}[/itex]

From the conclusion of all three sets of equations, I have 5 unknowns, with only three equations
I figure from here I need to some kinematics:
By circular motion, the y'[itex]_{A}[/itex] = R*w[itex]_{f}[/itex], which helps eliminate one unknown
In addition, there is the length of rope connecting the three bodies, but considering the pulley has mass, I am not sure how to relate the velocities of point A and point B
I believe I get stuck a lot on the kinematic constraints, and here I am not sure how to solve the system of equations I developed

Any help is greatly appreciated
 
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  • #2
Seems to me this problem is solvable by elementary energy conservation principles.
Were you explicitly asked to analyze in terms of impulse momentum considerations?
 
  • #3
Yes, the problem is an exercise in using impulse momentum. I know there are other ways to solve it, but I was hoping to clear up some confusion I have with using impulse momentum.
 

1. What is the difference between impulse and momentum?

Impulse refers to the change in an object's momentum, while momentum is a measure of an object's motion and is equal to its mass times its velocity.

2. How is impulse related to force?

Impulse is directly proportional to force and the time over which the force is applied. This can be represented by the equation: Impulse = Force x Time.

3. Can the impulse-momentum theorem be applied to all types of collisions?

Yes, the impulse-momentum theorem can be applied to both elastic and inelastic collisions. However, in elastic collisions, the total momentum of the system is conserved, while in inelastic collisions, some of the kinetic energy is lost.

4. How does the impulse-momentum theorem relate to Newton's second law?

The impulse-momentum theorem is essentially an extension of Newton's second law. It states that the change in an object's momentum is equal to the net force acting on it multiplied by the time it is applied.

5. Can momentum be negative?

Yes, momentum can be negative. This usually occurs when an object is moving in the opposite direction of a chosen positive direction. In this case, the momentum would be represented by a negative value but still follows the same laws and principles as positive momentum.

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