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Question

A rotating ball hits a rough horizontal plane with a vertical velocity $v$ and angular velocity $w$ . Given that the coefficient of friction is $u$ and the vertical component of the velocity after the collision is $v / 2$ , find:

a) the angular velocity after collision;

b) the impulsive ground reaction during the collision

Joshi sir comment

on applying momentum impulse theorem along vertical

-mv+Ndt = mv/2 (1)

by angular momentum angular impulse theorem

$2 m ω r^{2} / 5 - μ N d t r = 2 m ω^{,} r^{2} / 5 (2)$ By eq. (2) get angular velocity.

impulsive ground reaction = $\sqrt{{(N d t)}^{2} + {(μ N d t)}^{2}}$ by using eq. (1), solve it

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