Maximizing Vertical Height in Frictionless Rolling Motion on an Incline

AI Thread Summary
A hollow sphere of mass M and radius R rolls down a frictional incline and reaches point O before continuing onto a frictionless curve. The conservation of energy principle is applied, where potential energy at height h is converted into kinetic energy as it rolls. The discussion emphasizes that the sphere's rotation contributes to its kinetic energy, which remains constant as it transitions to the frictionless section. The key equation used is mgh = 1/2 M W² + 1/2 MV², indicating that the height reached after point O can be determined by this energy conservation. The problem is resolved, confirming that the sphere's rotation does not change upon entering the frictionless curve.
John Kolby
Messages
3
Reaction score
0

Homework Statement


A hollow sphere of mass M and radius R (I = 2MR2 /3) is released from rest at height h and rolls down a curved surface without slipping until it reaches the lowest point, O..

The curve to the right of O is frictionless. If the sphere continues past point O, what vertical height will it reach?
(In the diagram, the curve looks like a semi circle)

Homework Equations


KEr = 1/2 M W2
KEl = 1/2MV2

The Attempt at a Solution


According to my knowledge, if its frictionless and energy is conserved it should reach the same high.

mgh = 1/2 M W2 + 1/2MV2
**If I did something wrong I apologize, first time on this forum
 
Physics news on Phys.org
John Kolby said:

Homework Statement


A hollow sphere of mass M and radius R (I = 2MR2 /3) is released from rest at height h and rolls down a curved surface without slipping until it reaches the lowest point, O..

The curve to the right of O is frictionless. If the sphere continues past point O, what vertical height will it reach?
(In the diagram, the curve looks like a semi circle)

Homework Equations


KEr = 1/2 M W2
KEl = 1/2MV2

The Attempt at a Solution


According to my knowledge, if its frictionless and energy is conserved it should reach the same high.

mgh = 1/2 M W2 + 1/2MV2

**If I did something wrong I apologize, first time on this forum
Hello John Kolby. Welcome to PF !

Does the rotation of the sphere change once it gets to the frictionless part of the curve?

Does its rotation correspond to some amount of kinetic energy?
 
SammyS said:
Does the rotation of the sphere change once it gets to the frictionless part of the curve?

Does its rotation correspond to some amount of kinetic energy?
I don't think the rotation changes. The rotation accounts to Kr which I account for with 1/2MW2
 
Solved it thank you!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top