Question

A cylinder of mass m and radius R is rolling without slipping on a horizontal surface with angular velocity  ω0 . The velocity of centre of mass of cylinder is  ω0 R . The cylinder comes across a step of height R/4 . Then the angular velocity of cylinder just after the collision is (Assume cylinder remains in contact and no slipping occurs on the edge of the step)

Answer 1

sorry for blurry pictures....by using the concept of conservation of angular momentum about point p.....ang the torque of mg will be be not taken in this case because mg is not a impulsive force....

Answer 2

Answer:

The angular velocity after collision is $\dfrac{5}{6}\omega_{0}$.

Explanation:

Given that,

Mass of cylinder = m

Radius of cylinder = R

Angular velocity=$\omega_{0}$

Velocity of center of mass = $\omega_{0}R$

Height $h=\dfrac{R}{4}$

Using conservation of angular momentum

Initial angular momentum = final angular momentum

The angular momentum is the product of the moment of inertia and angular velocity.

$L=r_{cm}mv_{cm}+I_{cm}\omega$

$\dfrac{3}{2}mR^2\omega=(R-\dfrac{R}{4})mv_{0}+\dfrac{1}{2}mR^2(\dfrac{v_{0}}{R})$

$\dfrac{3}{2}R\omega=\dfrac{3}{4}v_{0}+\dfrac{1}{2}v_{0}$

$\omega=\dfrac{5}{6}\times\dfrac{v_{0}}{R}$

$\omega=\dfrac{5}{6}\omega_{0}$

Hence, The angular velocity after collision is $\dfrac{5}{6}\omega_{0}$.