Question

A uniform rod of mass m and length l is placed horizontally on a smooth horizontal surface an impulse p is applie to end b the time taken by the rod?

Answer 1

Let v is the velocity of particle of mass m just before collision.
$v_1$ is the velocity of centre of mass of rod and $\omega_1$ is the angular velocity of its centre just after collision.

using conservation of linear momentum,
$mv=Mv_1$
$v_1=\frac{mv}{M}$

from conservation of angular momentum,
$m\left(v\frac{L}{2}\right)=I\omega_1$

I is moment of inertia of rod about its centre of mass . so, I = 1/12 mL²

mvL/2 = 1/12mL² $\omega_1$

$\omega_1$ = 6mv/ML = 6P/ML

Let T is the time taken by rod to turn through right angle , $\frac{\pi}{2}$
so, T = π/2$\omega_1$
= π/2{6P/ML}
=πML/12P

Answer 2

Let v is the velocity of particle of mass m just before collision.

is the velocity of centre of mass of rod and  is the angular velocity of its centre just after collision.

using conservation of linear momentum,

from conservation of angular momentum,

I is moment of inertia of rod about its centre of mass . so, I = 1/12 mL²

mvL/2 = 1/12mL²  

= 6mv/ML = 6P/ML

Let T is the time taken by rod to turn through right angle ,  

so, T = π/2

= π/2{6P/ML}

=πML/12P