Acceleration of Pulley w/ Masses: Solving Homework

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Homework Help Overview

The problem involves a pulley system with two different masses and a solid cylinder pulley. The original poster seeks to determine the acceleration of the masses when the pulley is frictionless and the average frictional torque when one mass is given an initial speed and comes to rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply equations of motion and rotational dynamics but encounters difficulty due to multiple unknowns. Some participants suggest using the relationship between angular acceleration and linear acceleration to simplify the problem.

Discussion Status

Participants are actively engaging with the problem, providing insights on the relationships between variables. There is a recognition of the need to clarify the connection between angular and linear acceleration, which may help in resolving the original poster's confusion.

Contextual Notes

The discussion includes the challenge of managing multiple unknowns in the equations and the importance of recognizing fundamental relationships in mechanics. No explicit consensus has been reached, but guidance has been offered to aid understanding.

jmcmillian
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Homework Statement


A string passing over a pulley has a 3.8-kg mass hanging from one end and a 3.15-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.0 cm and mass .80 kg.
(a) If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
(b) In fact, it is found that if the heavier mass is given a downward speed of .2 m/s, it comes to rest in 6.2 s. What is the average frictional torque acting on the pulley?


Homework Equations


I= .5*M*R^2
For mass 1: F_1 - (m_1g) = m_1a
For mass 2: m_2g - F_2 = m_2a


The Attempt at a Solution


The inertia for the pulley is simple. .5*.8*.04^2 = .00064

Now, rearranging the first two equations:
a= (F_1 - m_1g)/(m_1)
a=(m_2g-F_2)/(m_2)

Since the magnitude of the acceleration has to be the same for both masses...

And now is where I am stuck. I have 3 unknowns (F_1, F_2, a), and the Inertia equation times alpha can be equivalent to (F_2 - F_1)*r. But, that introduces alpha into the mix, which would be another unknown.

What am I missing? Any help is greatly appreciated!
 
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I believe you can simply make this problem easier by elimitating one unknown either a, or alpha since there is a relationship between the two:

a = alpha * r (Not only the linear acceleration of the two masses will have the same magnitude but the one of the pulley as well since they are all in the same system)
By using this equation should be a little easier to find the acceleration since you would have three equations and three unknowns. (a, F2 and F1) and you would get another eqatuion : a = (F2 - F1)r^2/I in addition to the ones you already mentioned

I hope this helps!
 
You are missing the simple relationship between alpha and the linear acceleration.
 
Thank you both Doc Al and Hells Kitchen. The relationship between the alpha and linear accel. was floating around in my mind during my post, but wasn't really certain how to go about it. Sometimes I get so sucked into the problem that I lose focus and forget simple relationships that are the keys to success. Thanks again!
 

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