Question

A uniform rectangular plate of mass m which is free to rotate about the smooth vertical hinge passing through the center and perpendicular to the plate is lying on a smooth horizontal surface. A particle of mass m moving with speed 'u' collides with the plate and sticks to it as shown in the figure. The angular velocity of the plate after the collision will be

• A. $\frac{12u}{5a}$
• B. $\frac{12u}{19a}$
• C. $\frac{3u}{2a}$
• D. $\frac{3u}{5a}$

Answer 1

The correct option is D $\frac{3u}{5a}$

Angular momentum of the system above hinge should be.

Angular momentum of particle of mass $m$ before collision is,

$L_p=a.mu$

$L_p=mua$

Let after collision the angular velocity of the attached system be $\omega$

Moment of inertia about a perpendicular axis passing through hinge is

$I=$ moment of inertia of particle of mass $m$ about hinge + Moment of inertia of the plate about hinge.

$m.\left(\frac{\sqrt{5a}}{2}\right)+\frac{1}{12}m(a^2+(2a{)}^2)$

$=\frac{5}{4}m{a}^2+\frac{5}{12}m{a}^2$

$=\frac{20}{12}m{a}^2$

Applying conservation of angular mamentum we get

$I_w=L_p$

$=\frac{5}{3}m{a}^2$

$\frac{5}{3}m{a}^2.\omega =mua$

$\Rightarrow \frac{3u}{5a}$