Rotational Energy and Pulley System

AI Thread Summary
The discussion revolves around solving a physics problem involving a pulley system and rotational energy. The initial calculations for the speed of a 4.00-kg block were incorrect due to overlooking the kinetic energy contributions of both blocks. After incorporating the kinetic energy into the energy conservation equation, the participant initially calculated a speed of 5.71 m/s but later corrected it to 2.81 m/s after identifying a numerical error. The conversation highlights the importance of careful calculations and the clarity of problem presentation. The final consensus confirms the correct approach to the problem, emphasizing the value of collaborative troubleshooting in physics homework.
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Homework Statement


Problem reads:
The pulley in the figure has radius 0.160 m and a moment of inertia 0.480 kgm^2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.
YF-09-36.jpg

Homework Equations


E=\frac{1}{2}I\omega^{2}+m_{1}gh_{1}+m_{2}gh_{2}

E_{initial}=E_{final}

v=\omega r

The Attempt at a Solution


I said m_{1} was the 4.00 kg block and m_{2} was the 2.00 kg block. Setting the initial and final energies of the system equal, I got:

m_{1}gh_{10}+m_{2}gh_{20}+\frac{1}{2}I\omega^{2}_{0}=m_{1}gh_{1}+m_{2}gh_{2}+\frac{1}{2}I\omega^{2}

Since initial angular momentum is zero and so is the height of block 2, and in the final state, height of block 2 is zero, this simplifies to:

m_{1}gh_{10}=m_{2}gh_{2}+\frac{1}{2}I\omega^{2}

Plugging in numbers:
(4.00)(9.8)(5.00)=(2.00)(9.8)(5.00)+\frac{1}{2}(0.480)\omega^{2}

I got \omega=20.21 rad/s. Then using v=\omega r I just plugged in the radius and the angular velocity I just found to get v=3.23 m/s

This is not the right answer according to the homework website unfortunately... help would be appreciated!
 
Last edited:
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You are forgetting that not all of the kinetic energy is in the pulley. The blocks have some too.
 
Alright, I added the kinetic energy into the equation (kept the left side the same since kinetic energy is 0 for both blocks when they're at rest) and solved for v:

m_{1}gh_{10}=m_{2}gh_{2}+\frac{1}{2}I\omega^{2}+\frac{1}{2}m_{1}v^{2}+\frac{1}{2}m_{2}v^{2}

Using v=\omega r and isolating \omega and substituting that into the equation, I got 5.71 m/s.

But it still says this is wrong... how do I go about getting the right answer?
 
You are doing everything right and your presentation of this problem is super clear, good job. But if I put the numbers into your equation I don't get what you get for v. Are you just punching in numbers wrong?
 
Ah alright, I got it (v=2.81 m/s)... it turns out I did punch in a wrong number somewhere, I hate it when I do that, heh.

Anyway, thank you for the help Dick, I really appreciate it!
 
That's what I get as well. Like I say, I wish all posters presented stuff this well.
 
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