- #1
Reshma
- 749
- 6
A solid sphere of mass m and radius 'a' is rolling with a linear speed v on a flat surface without slipping. The magnitude of the angular momentum of the sphere with respect to a point along the path of the sphere on the surface is?
Moment of inertia of the sphere along any of its diameters is:[itex]{2\over 5}ma^2[/itex]
By parallel axis theorem:
Moment of inertia of the sphere rolling with respect to a point on the surface is:[itex]{2\over 5}ma^2 + ma^2 = {7\over 5}ma^2[/itex]
So the angular momentum will be:[itex]L = {7\over 5}ma^2\omega ={7\over 5}mav[/itex]
The options given to me are:
[itex]{2\over 5}mav[/itex]
[itex]{7\over 5}mav[/tex]
[itex]mav[/itex]
[itex]{3\over 2}mav[/itex]
Is my answer correct?
Moment of inertia of the sphere along any of its diameters is:[itex]{2\over 5}ma^2[/itex]
By parallel axis theorem:
Moment of inertia of the sphere rolling with respect to a point on the surface is:[itex]{2\over 5}ma^2 + ma^2 = {7\over 5}ma^2[/itex]
So the angular momentum will be:[itex]L = {7\over 5}ma^2\omega ={7\over 5}mav[/itex]
The options given to me are:
[itex]{2\over 5}mav[/itex]
[itex]{7\over 5}mav[/tex]
[itex]mav[/itex]
[itex]{3\over 2}mav[/itex]
Is my answer correct?