- #1
Milo Martian
- 18
- 0
1. The problem statement:
Consider a solid sphere and a hollow sphere, of equal mass M and equal radius R ,at rest on top of an incline . If there is no slipping which will reach the bottom faster.
2. Homework Equations :
acm = Fext/M (cm= centre of mass)
angular acceleration= torqueext/ I ( I = moment of inertia)
Ihollow sphere= 2MR2/3
Isolid sphere=2MR2/5
3. The Attempt at a Solution :
acm should be same for both of them as Fext and M are same, so they should reach at same time. But angular acceleration of solid sphere will be more since Isolid sphere< Ihollow sphereand torque is same for both, so by this info the solid sphere should reach faster. Thus, the confusion.Also, how can acm remain same for both when angular acceleration of solid sphere is higher , the acm of the solid cylinder should be higher since the radius is same for both. But according to the first equation, acm should be same.
Consider a solid sphere and a hollow sphere, of equal mass M and equal radius R ,at rest on top of an incline . If there is no slipping which will reach the bottom faster.
2. Homework Equations :
acm = Fext/M (cm= centre of mass)
angular acceleration= torqueext/ I ( I = moment of inertia)
Ihollow sphere= 2MR2/3
Isolid sphere=2MR2/5
3. The Attempt at a Solution :
acm should be same for both of them as Fext and M are same, so they should reach at same time. But angular acceleration of solid sphere will be more since Isolid sphere< Ihollow sphereand torque is same for both, so by this info the solid sphere should reach faster. Thus, the confusion.Also, how can acm remain same for both when angular acceleration of solid sphere is higher , the acm of the solid cylinder should be higher since the radius is same for both. But according to the first equation, acm should be same.