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A circular wooden loop of mass m and radius R rest on a flat horizontal friction less surface. A bullet, also of mass m, and moving with a velocity V, strikes the loop and gets embedded in it. The thickness of the loop is much smaller than R. The angular velocity with which the system angular velocity with which the system rotates just after the bullet strikes the loop is

 

\[\begin {array} {1 1} (a)\;\frac{v}{4}R & \quad (b)\;\frac{v}{3}R \\ (c)\;\frac{2v}{3}R & \quad  (d)\;\frac{3v}{4R} \end {array}\]
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centre of mass of the loop - bullet system is $R/2$ from centre of loop.
By momentum conservation :
$mv=2m V_{cm}$
$V_{cm}=\large\frac{v}{2}$
Applying angular momentum conservation about the centre of mass.
$MVR/2=I_cw$
$I_c$ for ring $= mr^2+m(r/2)^2$
$I_c$ of bullet $= m (r/2)^2$
moment of inertia of the system $I_c= \large\frac{6mR^2}{4}$
$\qquad= \large\frac{3}{2} m$$R^2$
$mv \large\frac{R}{2}= \frac{3}{2}$$ mR^2w$
$w= \large\frac{v}{3}$$R$
answered Dec 6, 2013 by meena.p
edited Jun 24, 2014 by lmohan717
 

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