Which will reach the bottom faster: solid or hollow sphere?

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SUMMARY

The discussion centers on the dynamics of a solid sphere versus a hollow sphere of equal mass (M) and radius (R) rolling down an incline without slipping. Key equations include the center of mass acceleration (acm = Fext/M) and the moments of inertia for the hollow sphere (Ihollow = 2MR²/3) and solid sphere (Isolid = 2MR²/5). The solid sphere accelerates faster due to its lower moment of inertia, leading to a higher angular acceleration despite both spheres experiencing the same net external force. The confusion arises from the relationship between translational and rotational motion in pure rolling scenarios.

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  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and torque
  • Knowledge of moments of inertia for different shapes
  • Concept of pure rolling motion and static friction
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  • #31
llober said:
I think is better to do an Energy balance...
https://www.physicsforums.com/attachments/104346
But won't mechanical energy change due to friction? just curious
 
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  • #32
Milo Martian said:
But won't mechanical energy change due to friction? just curious
If there's no sliding, there's no loss by friction, because the no sliding translates into speed zero at the contact point, so the "friction force" does not move.
 
  • #33
llober said:
If there's no sliding, there's no loss by friction, because the no sliding translates into speed zero at the contact point, so the "friction force" does not move.
OK
 
  • #34
Milo Martian said:
Okay, i got it .
Fext= mgsinA - f= macm ...(1)
fXR= torque = I X angular acc. = I X acm/R
=> f= Iacm/R2 ...(2)
From (1) and (2) :
macm = mgsinA - Iacm/R2
=> acm= mgsinA/(m + I/R2)
So, since Isolid sphere< Ihollow sphere
Therefore, acm for solid sphere> acm for hollow sphere.
i hope i am right. phew
You got it.
 

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