Recent content by Jnan

Okay now I understand it quite well. Let us now initiate a discussion about the linear momentum.

No force would produce torque about that. I think any axis passing through the point of contact should work fine.

One which passes through both CM and point of contact. ..

Friction|||

Axis might be one which is vertical, passing through CM and along the plane of paper. Then normal and Mg will have zero torque.

Anyone who would like to add more about the angular momentum story..

OH yeah.. I agree with that.. So with the centre of mass we can conserve angular momentum right...

I didn't really get your statement. How can pure rolling have a effect on KE?

Okay this should be the point of contact of the ball with ground.

Yes it assumes that...

Oh sorry. That was a huge blunder from my part... I just considered that after freezing the angular speed and linear speed will be same as before (totally wrong) . If that would have been the case, angular momentum will be conserved from any point. Any more views about the topic? Thanks for...

Please help me out.. This is screwing up my head.

Homework Statement If we have a hollow ball completely filled with water which is rolling without slipping on a horizontal ground. If the water freezes which of the parameter will remain unchanged- angular speed, angular momentum, linear momentum, kinetic energy, total energy Homework...

Let us consider force as a function of time as F(t). We know, dP/dt = F(t) m dv/dt = F(t) m∫dv = ∫F⁽t⁾dt m⁽v₂-v₁) = θ (θ=∫F⁽t⁾dt with limits t₂ and t₁) Now, for a certain situation the initial velocity⁽v₁) and mass can be easily recorded. Also the time interval for which work...

The Coefficient of Restitution Is actually defined as such. It is actually the ratio of impact after and before collision. Since most of the times, we don't consider loss of mass of the colliding bodies we have only velocity as the parameter which will measure the impact of the collision...