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 the centre of the mass cylinder is${\omega }_{o}R$. The cylinder comes across a step of height$R/4$. Then the angular velocity of the cylinder just after the collision is (assume cylinder remains in contact and no slipping occurs on the edge of the step)

A

$\frac{5{\omega }_{0}}{6}$

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B

${\omega }_{0}$

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C

$2{\omega }_{0}$

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D

$6{\omega }_{0}$

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Solution

## The correct option is A $\frac{5{\omega }_{0}}{6}$Step 1: Given dataMass of cylinder$=m$Radius of cylinder$=R$Angular velocity of the cylinder before the collision$={\omega }_{0}$The velocity of the centre of mass, ${v}_{0}={\omega }_{0}R$Height, $h=\frac{R}{4}$The radius of the centre of mass, ${R}_{0}=R-\frac{R}{4}=\frac{3R}{4}$Step 2: AssumptionsInitial angular momentum$={L}_{i}$Final angular momentum$={L}_{f}$Velocity of the after collision$=v$Angular velocity of the cylinder after the collision$=\omega$Step 3: Calculation of the angular velocity of the cylinder just after the collisionThe cylinder remains in contact with the step after the collision and hence the collision is “in-elastic”.After the collision, torque about point “P” is zero as depicted by the diagram and hence the law of conservation of angular momentum can be applied.According to the basic crux of the law of conservation of linear momentum, “the initial angular momentum must be equal to the final angular momentum”.${L}_{i}={L}_{f}$ ………………………….(a) Step 3.1: Calculation of the initial angular momentumInitial angular momentum, ${L}_{i}=mv{R}_{0}+\frac{1}{2}m{R}^{2}{\omega }_{0}$${L}_{i}=m{v}_{0}\left(\frac{3R}{4}\right)+\frac{1}{2}m{R}^{2}{\omega }_{0}$ ${L}_{i}=\frac{3}{4}m{v}_{0}R+\frac{1}{2}m{R}^{2}{\omega }_{0}$ ………………………………(b) Step 3.2: Calculation of the final angular momentum and the angular velocity of the cylinder after collisionFinal angular momentum, ${L}_{f}=\frac{1}{2}m{R}^{2}\omega +m{R}^{2}\omega$${L}_{f}=\frac{3}{2}m{R}^{2}\omega$……………………….(c)Using equations (b) and (c) in equation (a), we get$\frac{3}{2}m{R}^{2}\omega =\frac{3}{4}m{v}_{0}R+\frac{1}{2}m{R}^{2}{\omega }_{0}$Using, ${\omega }_{0}=\frac{{v}_{0}}{R}$$\frac{3}{2}m{R}^{2}\omega =\frac{3}{4}m{v}_{0}R+\frac{1}{2}m{R}^{2}\left(\frac{{v}_{0}}{R}\right)\phantom{\rule{0ex}{0ex}}\frac{3}{2}R\omega =\frac{3}{4}{v}_{0}+\frac{1}{2}{v}_{0}\phantom{\rule{0ex}{0ex}}\frac{3}{2}R\omega =\frac{3{v}_{0}+2{v}_{0}}{4}\phantom{\rule{0ex}{0ex}}\frac{3}{2}R\omega =\frac{5{v}_{0}}{4}\phantom{\rule{0ex}{0ex}}\omega =\frac{5}{6}×\frac{{v}_{0}}{R}\phantom{\rule{0ex}{0ex}}\omega =\frac{5}{6}{\omega }_{0}$Hence, option A is correct.

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