Question

A disc of mass $M$ radius $R$ is fixed about an axis as shown in figure. If a particle of mass $\frac{M}{4}$ moving with speed $v$ perpendicular to the disc plane, hits the disc at point $P$ and sticks to the disc, the disc starts rotating about its axis as shown. Find the angular velocity of the system (disc + particle) after the hit.

• A. $\omega =\frac{v}{R}$
• B. $\omega =\frac{v}{5R}$
• C. $\omega =\frac{v}{2R}$
• D. $\omega =\frac{v}{\sqrt{5}R}$

Answer 1

The correct option is C $\omega =\frac{v}{2R}$

As we know from angular momentum conservation.
$L=mv{r}_{\perp }=I\omega$
MOI of the disc about an axis passing through its centre and perpendicular to its plane is $I=\frac{M{R}^2}{2}$
Hence by perpendicular axis theorem, MOI of the disc about given axis,
$I^{\prime }=\frac{M{R}^2}{4}$
where $I^{\prime }$ is the MOI about an axis passing through disc centre and in the plane of disc.
Initial angular momentum $=mv{r}_{\perp }$
$L_i=\frac{M}{4}.v.R$
Final angular momentum of the system
$L_f=I\omega$
$L_f=(\frac{M{R}^2}{4}+\left(\frac{M}{4}\right)R^2)\omega$
$L_f=\frac{M{R}^2}{2}\omega$
Hence, from angular momentum conservation
$L_i=L_f$
$\frac{MvR}{4}=\left(\frac{M{R}^2}{2}\right)\omega$
i.e $\omega =\frac{v}{2R}$