Determine Mass of Pulley in Atwood's Machine

AI Thread Summary
To determine the mass of the pulley in an Atwood's machine with masses m1 = 5.0 kg and m2 = 3.0 kg, the initial and final velocities of m1 are provided, along with its height. The acceleration of m1 is calculated to be 2.16 m/s², and it is confirmed that the final velocity of m2 is also 1.8 m/s. The relationship between the linear acceleration of the masses and the angular acceleration of the pulley is established, with the latter being 18 rad/s². The net torque on the pulley can be expressed as τ = Iα, leading to the calculation of the pulley's mass. The discussion emphasizes the importance of understanding the dynamics of the system to solve for the mass effectively.
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Homework Statement


The two masses(m1 = 5.0 kg and m2 = 3.0 kg) in the Atwood's machine shown below are released from rest, with m1 at a height of 0.75 m above the floor. When m1 hits the ground its speed is 1.8 m/s. Assuming that the pulley is a uniform disk with a radius of 12 cm, determine the pulley's mass.

http://img23.imageshack.us/img23/904/pulleyl.jpg​
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Homework Equations


Rotational kinematics


The Attempt at a Solution


I don't really know where to start here. Could someone please start me off?
 
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Initial and final velocity of m1 is given. h is given. Find acceleration of m1. m2 and pulley will have the same acceleration. Write down the expression for a to them. Then solve for the mass of the pulley.
 
rl.bhat said:
Initial and final velocity of m1 is given. h is given. Find acceleration of m1. m2 and pulley will have the same acceleration. Write down the expression for a to them. Then solve for the mass of the pulley.

Is the final velocity of m2 also 1.8 m/s? And is the mass of the pulley simply the mass of the disk?
 
Precursor said:
Is the final velocity of m2 also 1.8 m/s? And is the mass of the pulley simply the mass of the disk?
Yes.
 
So I found the acceleration to be 2.16 m/s^2.

Now is the acceleration of the pulley 2.16/0.12 = 18 rad/s^2?

So how do I find the mass?
 
Alright! You are almost there. You can say that net torque on the disk is \tau=I\alpha
 
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