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A uniform circular disc of radius r is placed on a rough horizontal surface and given a linear $w_0$ as shown. The disc com,es to rest after moving some distance to the right. It follows that

\[\begin {array} {1 1} (a)\;3V_0=2 w_0 r & \quad (b)\;2V_0=W_0 r \\ (c)\;\frac{W}{2} & \quad  (d)\;none \;of\;these \end {array}\]

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Considering bottom most point , $V_p= V_0+rw_0$
If disc stops, both V and w become zero.
Angular momentum about P is conserved.
$L_i=mv_0r-I_0w_0=L_{final}=0$
$\therefore mv_0 r =I_0w_0$
$mv_0r= \large\frac{mr^2}{2}w_0$
$\therefore 2V_0 =r w_0$
answered Dec 5, 2013 by meena.p
 

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