Did I Calculate the Velocity of a Cylinder Correctly Using Energy?

AI Thread Summary
The discussion centers on calculating the velocity of a cylinder using energy conservation principles. The original poster struggles with the equations related to gravitational potential energy and kinetic energy, noting discrepancies with the answer key. A suggestion is made to reframe the equations in terms of total energy at the beginning versus the total energy at the top, which helps identify a sign error in the calculations. The final resolution confirms that the equations were correct but needed to be expressed in terms of energy gain rather than loss. The poster expresses gratitude for the assistance and plans to apply the suggested method in future calculations.
jisbon
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Homework Statement
Cylinder with mass 1.5kg and 0.1m radius roll without slipping up a slope of length 3m and height 1m. Cylinder has an initial translational velocity of 4m/s. Calculate velocity of the cylinder as it leaves the top of the slope.
Relevant Equations
Gain in GPE = Loss in translational KE + loss in rotational KE
I seem to be able to do this problem (at least from what I think, but my answer is still wrong according to the answer key, please do help check.)

Since:
Gain in GPE = Loss in translational KE + loss in rotational KE
##\left(m\cdot g\cdot h\right)=\left(\frac{1}{2}\left(m\right)\left(v_{f}^2\right)-\frac{1}{2}\left(m\right)\left(v_{i}^2\right)\right)+ \left(\frac{1}{2}\left(I\right)\left(\omega_{f}^2\right)-\frac{1}{2}\left(I\right)\left(\omega_{i}^2\right)\right)##
Whereby
##I = 0.5 (mr^2) = 0.5 (1.5*0.1^2)##
##\omega = v/r = 4/0.1##

##\left(1.5\cdot 9.8\cdot 1\right)=\left(\frac{1}{2}\left(1.5\right)\left(v_{f}^2\right)-\frac{1}{2}\left(1.5\right)\left(4^2\right)\right)+\left(\left(\frac{1}{2}\left(0.5\cdot \:1.5\cdot \:0.1^2\right)\left(\frac{v_{f}}{0.1}\right)^2\right)\:-\:\left(\frac{1}{2}\left(0.5\cdot 1.5\cdot 0.1^2\right)\left(\frac{4}{0.1}\right)^2\right)\right)##

Are the equations correct/did I miss something out? Thanks
 
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Since it does not slip, what happens to rotational speed in relation to translational speed? Ok, i think you took care of that. I'm on my phone so it's hard to read all of the symbols.
 
I think I see it now. You really have what appears to be just an error of swapping the signs. Try writing it like this: (Total energy at beginning) = (Total energy at top) to see how that works, and if you spot the error.
 
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scottdave said:
I think I see it now. You really have what appears to be just an error of swapping the signs. Try writing it like this: (Total energy at beginning) = (Total energy at top) to see how that works, and if you spot the error.
Solved the error. Thanks for your help. Will use your method in the future
 
jisbon said:
Solved the error. Thanks for your help. Will use your method in the future
Either method is fine. The problem was that your reference equation was in terms of loss of energy, but you turned that into final - initial, which would be a gain in energy.
It is generally safer to record reference equations in terms of gain.
 
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