Rotational Motion & Energy Equation (Intro Physics)

AI Thread Summary
The discussion focuses on applying the energy equation to a problem involving kinetic energy, potential energy, and rotational motion. The initial energy is entirely potential due to the mass M, which converts to kinetic energy as it falls. It's emphasized that the velocity v of the mass must be included in the calculations, specifically in the kinetic energy term (1/2)Mv^2. The conversation confirms that all three energy terms—kinetic energy of the falling mass, kinetic energy of the turntable, and initial potential energy—should be accounted for, assuming no energy loss due to friction. Overall, the key takeaway is the importance of energy conservation in solving the problem.
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Homework Statement



[PLAIN]http://carlodm.com/phys/1.png

Homework Equations



Energy Equation

The Attempt at a Solution



As shown in the scan.

Is this the correct approach? I have a feeling that the velocity, v, needs to be used. I used the Energy Equation for this problem but wasn't sure if it was the right equation to address kinetic energy, potential energy, and rotational motion.

Please lead me to the right direction if I'm incorrect.

Thanks
 
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This assumes there is no energy lost due to friction, of course.
The initial energy is all potential in the mass M
At the end, the potential energy lost by the falling mass (Mgh) has appeared as kinetic energy in both the mass and the rotating table.
So you do need to consider the velocity v of the mass.
 
Hi StoneBridge,

Thanks for your response. How do I account for the kinetic energy of the linear 'falling of the mass' when I am only given the velocity?

For the right hand side of the equation, do I just include another kinetic energy term (... KE = (1/2)Mv^2)

Would that account for the velocity?

Thanks
 
The question just asks that you express everything in terms of what's given in the question. You have the mass M and the velocity v of the mass as given.
So yes, ½Mv2 is the k.e. of the falling mass, and in your notes I see the expression for the k.e. of the turntable and the p.e. the mass has at the start.
So it's just a case of including all 3 terms and remembering that energy is conserved (no friction here!)
 
Thanks a lot! I appreciate it!
 
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