Finding Kinetic Energy and Velocity of a pulley / block system

AI Thread Summary
The discussion focuses on calculating the kinetic energy and velocity of a pulley-block system where a 1kg block is attached to a 75kg pulley. The potential energy of the block is converted into kinetic energy as it falls from a height of 50cm. The moment of inertia for the pulley is identified as I = 1/2(M)(R^2), but the user struggles with incorporating this into the kinetic energy equation due to the presence of two unknowns: velocity and angular velocity. Clarification is sought on how the hanging mass influences the moment of inertia and how to solve for the kinetic energy effectively. Understanding the relationship between linear and angular motion is crucial for solving the problem.
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Homework Statement


In the problem, a hanging block, m, is attached to a pulley that is mounted on a horizontal axle with negligible friction. The pulley has a mass of M=75kg and a radius of R=6.0cm. The mass of the hanging block is m=1kg. The block is released from rest at a height above the ground of 50cm without the cord slipping on the pulley. The pulley is a uniform disk.


Homework Equations


A) What is the Kinetic Energy of the system right before the block hits the ground?
B) What is the speed of the block right before it hits the ground?

The Attempt at a Solution


I understand that the potential energy of the system in the beginning will be equal to the kinetic energy of the system just before the block hits the ground, however, I do not know how to find the moment of inertia of the system or the velocity of the block.

I have this equation setup:
PE = KE
mgh = 1/2(m)(v^2) + 1/2(I)(ω^2)

I am not sure where to even start since I don't know how to find the moment of inertia, I, or omega. Anything would help out, even a little nudge in the right direction would be appreciated.
 
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Moment of inertia of pulley = moment of inertia of a uniform disc = \frac{1}{2}Mr^{2}.
 
grzz said:
Moment of inertia of pulley = moment of inertia of a uniform disc = \frac{1}{2}Mr^{2}.

Would the hanging mass affect the moment of inertia on the disc?
 
I understand that the moment of inertia for the disc is 1/2(M)(r)^2, but how do you solve the equation for KE? There are two unknown variables still in the equation, and I am not sure how to solve for velocity or omega.
 
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