# Collision of Rolling body

When a rolling body collides with a wall elastically,its torque is zero.But direction of velocity of particle is reversed.So will the angular momentum be conserved?

• I think it should remain conserved ,in intervals,before the ball.colloide,during ball.collide,after collision ,and direction angular momentum is along angular velocity vector,which has changed after collision – Yuvraj Singh... Sep 15 '19 at 11:54
• But mvr direction changes then how angular momentum be conserved – Muhammed Yoosuf Sep 15 '19 at 11:55
• Angular velocity direction is same after collision but direction of v is changed then how in this situation angular momentum be conserved – Muhammed Yoosuf Sep 15 '19 at 11:57
• Yes angular momentum will be conserved torque delta t=change in angular momentum since elastic collision k.e will be conserved, hence angular momentum will be conserved – Yuvraj Singh... Sep 15 '19 at 13:07

If there is no friction between the body and the ground then the body will rebound with the same speed as it had before hitting the wall and will rotate with an unchanged angular velocity.
Even though there will be relative motion between the body and the ground as there is no friction no mechanical energy is dissipated at the point of contact so such motion continues without change.

If there is friction between the body and the ground then the frictional force on the body will be in such a direction as to reduce the velocity of the centre of mass of the body whilst also providing a torque which will reduce the angular velocity of the body and eventually reverse the direction of rotation such that the no slip condition is satisfied.
During this time the body loses mechanical energy and heat is produced.

—-

There are two components to the angular momentum of the body.
The first component is due to the rotation of the body about its centre of mass and it is sometimes called the spin angular momentum.
The spin angular momentum does not change during the collision.

The second component is due the translational motion of the centre of mass and this is sometimes called the orbital angular momentum.
This is the angular momentum whose value is $$mvr$$ about a point on the ground.
The the body hits the wall it experiences a horizontal force due to the wall which reverses the direction of the linear momentum $$mv$$.
That force exerted by the wall can also be thought of a torque about a point on the ground acting on the body.
This torque reverses the direction of the orbital angular momentum $$mvr$$ about a point on the ground.

• Here friction is absent.But my doubt is that since direction of velocity reverses after collision wilk the direction of mvr changes? – Muhammed Yoosuf Sep 15 '19 at 12:10
• @MuhammedYoosuf The angular momentum cannot change because, in the absence of friction, there is no torque applied to body. – Farcher Sep 15 '19 at 12:12
• My doubt is that when velocity reverses after collision mvr will change to -mvr? Is it wrong or correct – Muhammed Yoosuf Sep 15 '19 at 14:29
• @MuhammedYoosuf The body will still be spinning in the same direction and with the same magnitude as before the collision. – Farcher Sep 15 '19 at 14:31
• Will the angular momentum produced due to translation to be conserved? – Muhammed Yoosuf Sep 15 '19 at 17:08

Basic definition for angular momentum to be conserved,there should be absence of sudden external impulsive force which can change the momentum,since there is no impulsive torque before ,during,after collision,since it is elastic collision hence k.e energy remain conserved hence velocity,remain same ,hence after all these argument ,it clearly describe that in given situation angular momentum remain conserved.

Linear moment isn't conserved when hitting a wall because the wall can't move and thus its momentum (which is zero) can't change.

Conversely, angular momentum isn't conserved, because angular momentum is just linear momentum at a distance. The interpretation of angular momentum conservation is that of conservation of the geometry (line in space) where linear momentum acts through.

In this case, before the impact, the center of rotation is at the contact point $$R$$ below the center of mass, and the momentum axis at a distance $$\tfrac{I_c}{m r}$$ above the center of mass (here $$I_c$$ is the mass moment of inertia of the object).

What happens next depends on if friction is considered or not. In the frictionless case, the ball is no longer rolling after the bounce. Its center of rotation is now $$R$$ above the center of mass, and the momentum axis $$\tfrac{I_c}{m r}$$ below the center of mass.

With friction, that situation is by far more complex, but suffice to say the angular momentum isn't conserved either.

There is a special coefficient of friction value of $$\mu = \tfrac{I_c}{m r^2}$$ which makes the ball instantly roll after the bounce, although it lifts up from the floor as friction causes a vertical impulse in addition to the regular bounce.