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Finding Rotational Kinetic Energy of Sphere on Ramp

  • Thread starter joel amos
  • Start date
1. The problem statement, all variables and given/known data
If Inga, the Laboratory Assistant, rolls a spare head down a 4 m ramp because it was spherical and solid and too heavy at 4.5 kg at a speed of 4.5 m/s, what was its total kinetic energy?


2. Relevant equations
KE = (1/2) I ω^2
I = (2/5) MR^2

3. The attempt at a solution
Basically, I'm not sure how to solve this without the sphere's radius. Any help, solutions, or hints would be highly appreciated.
 

haruspex

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Just go ahead with the algebra and you should find the radius cancels out. The rotational KE of a sphere is in fixed ratio to its linear KE, regardless of radius.
 
So does this look correct:

KE = (1/2)(4.5 kg)(4.5m/s)^2 + (1/2)(2/5)(4.5)r^2[(4.5m/s)/r]^2
KE = 6.75 J + 18.225 J
KE = 24.98 J
 
115
3
Where'd that 6.75 J come from?
 
Where'd that 6.75 J come from?
Hmm..I too am now wondering the same.

Is this better:

KE = 45.453 J + 18.225 J
KE = 63.69 J
 

haruspex

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Hmm..I too am now wondering the same.

Is this better:

KE = 45.453 J + 18.225 J
KE = 63.69 J
Still a bit low. I get 45.56+18.23=63.79
 
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Still a bit low. I get 45.56+18.23=63.79
That's also what I'm getting. Joel, I think you're not plugging into your calculator correctly.
 

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