Question

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A solid sphere of radius R and mass m rolls purely on rough horizontal surface. if it strikes the wall elastically then angular momentum of sphere just after strike is
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Answer 1

Hi !

So, I just found out the crux of the problem !

Let's consider the system to be both the sphere along with the wall.

Now,we have to find the angular momentum about the point of contact with the ground and though because we know that the sphere is rolling purely, we can conserve angular momentum about the point of contact considering the system to be ball and the wall together.

Now, so initial angular momentum = mv(R) + (2mR^2/5 + mR^2)w = mvR+7/5mR^2*v/R = mvR + 7/5mvR = 12/5mvR

So, this is the initial angular momentum of the system.

Note that I've calculated the moment of Inertia about the point chosen that is the point in contact with ground.

And also note that I've not put up any term of angular momentum of wall because obviously it's velocity and angular velocity is both null.

Now, coming to the amazing part.

Because the ball collides, both the ball and the wall would exert a normal force on each other but then if the ball was considered the system alone then this would have produced a torque on the ball due to normal force along the point of contact as chosen but then angular momentum would not have been conserved

So, if now that I have chosen both together as a system the net torque cancels out and so, now due to no external torque acting on the system , angular momentum remains conserved and also because the collision is elastic the velocity remains constant and thus same after collision.

So, because Initial angular momentum = Final angular momentum

Therefore, Final angular momentum that is tue angular momentum asked at the instant = 12/5 mvR

Hope this helps you !

Comment down if you've any doubts !

Answer 2

Answer:3/5mVR

Explanation: