How to solve rotational work and energy problem?

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SUMMARY

The discussion focuses on solving a rotational work and energy problem involving a basketball and a can of frozen juice rolling down a hill. The basketball reaches a translational speed of 6.12 m/s, leading to the calculation of the hill's height using the equation h = (1/2V^2 + 1/3(Tangent V)^2) / g, resulting in a height of 3.18 meters. The relationship between translational speed and angular velocity is clarified, confirming that tangent velocity is equivalent to translational velocity when rolling without slipping.

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Ailiniel
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Homework Statement



Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 6.12 m/s. Ignore frictional losses.

Homework Equations



(a) What is the height of the hill?

(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

The Attempt at a Solution



Initial Total mechanical energy = Final Total mechanical energy
1/2mv^2 + 1/2IW^2 + mgh = 1/2mv^2 + 1/2IW^2 + mgh
Initial KE = 0 and final PE = 0
mgh = 1/2mv^2 + 1/2IW^2
gh = 1/2v^2 + 1/2IW^2

I am stuck with 2 unknown variables W and h.
 
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Hint: Assuming each object rolls without slipping, what's the relationship between v and W?

Also: What's the rotational inertia of each object?
 
mgh = 1/2mv^2 + 1/2IW^2
Tangent Velocity = rW
W=Tangent V / r

Substitute W and solve for h.
h = (1/2V^2 + 1/3(Tangent V)^2) / g

Tangent V = translational V

h=3.18

I got it now, Thanks!
 
Last edited:
Is Tangent velocity the same as translational velocity?
Yes, because it rolls without slipping.
 
Ailiniel said:
I got it now, Thanks!
Good!
 

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