Question

A cylinder of mass m and radius R is rolling without slipping on a horizontal surface with angular velocity ${\omega }_0$. The velocity of center of mass cylinder is ${\omega }_{0}R$. The cylinder comes across a step of height $\frac{R}{4}$. 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)

• A. $\frac{5{\omega }_o}{6}$
• B. ${\omega }_0$
• C. $2{\omega }_o$
• D. $6{\omega }_0$

Answer 1

Answer: (A)

$\frac{5{\omega }_o}{6}$

Given,

Mass of cylinder $=m$

Radius of cylinder $=R$

Angular velocity $={\omega }_0$

Velocity of center of mass $={\omega }_{0}R$

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

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.

$L=r_{cm}m{v}_{cm}+I_{cm}\omega$

$\frac{3}{4}m{R}^2\omega =(R-\frac{R}{4})m{v}_0+\frac{1}{2}m{R}^2\left(\frac{v_0}{R}\right)$

$\frac{3}{4}R\omega =\frac{3}{4}{v}_0+\frac{1}{2}{v}_0$

$\omega =\frac{5}{6}\times \frac{v_0}{R}$

$\omega =\frac{5{\omega }_0}{6}$