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

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Homework Help Overview

The discussion revolves around a physics problem comparing the motion of a solid sphere and a hollow sphere of equal mass and radius rolling down an incline without slipping. Participants explore the implications of translational and rotational motion, particularly focusing on the relationship between angular acceleration and the acceleration of the center of mass.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion for both spheres, questioning how angular acceleration relates to the acceleration of the center of mass. There is confusion about whether the center of mass acceleration can be assumed equal for both spheres despite differences in their moments of inertia.

Discussion Status

The discussion is ongoing, with participants actively questioning assumptions and exploring the relationship between forces acting on the spheres. Some guidance has been offered regarding the net forces and torques involved, but no consensus has been reached on the implications for acceleration.

Contextual Notes

Participants note the importance of considering both translational and rotational motion, as well as the effects of friction and gravitational forces on the spheres' motion. There is an acknowledgment of the complexities introduced by the rigid body nature of the spheres and the conditions of pure rolling.

  • #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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