Question

A solid Sphere of Radius R and Mass m rolls purely on horizontal ground towards a smooth vertical wall. If it strikes the wall elastically then magnitude of angular momentum of sphere just after strike about the point of contact on the ground is:​

Answer 1

Answer:

final angular momentum = $\dfrac{2}{5}mv_0^2$

Explanation:

given,

* Radius of the sphere = R

* mass of the sphere = m

*initial velocity = $v_0$

* initial angular velocity = w

* moment of inertia of a sphere = $I = \dfrac{2}{5}mR^2$

* w = $\dfrac{v_0}{R}$

collision is elastic,

From the conservation of angular momentum,

final angular momentum = initial angular momentum

$P_f\ =\ Iw^2\ =\ \dfrac{2}{5}mR^2\times \left (\dfrac{v_0}{R}\ \right )^2\\\Rightarrow P_f\ =\ \dfrac{2}{5}mv_0^2$

Answer 2

Answer:3/5 mv0r

Explanation:

After the strike, the linear velocity is reversed and angular velocity remains unchanged.

I = mv0R – Icω

=mv0R-2/5mR^2(v0/R)

I = 3/5 mv0R