Question

A uniform thin rod of length $2L$ and mass m lies on a horizontal table. A horizontal impulse J is given to the rod at one end. There is no friction. The total K.E. of the rod just after the impulse will be.

• A. $J^2/2m$
• B. $J^2/m$
• C. $2J^2/m$
• D. $6J^2/m$

Answer 1

The correct option is C $2J^2/m$

The given impulse acts as both Linear and an Angular impulse.

Linear Impulse = $J=m{v}_{cm}$

Angular Impulse =$J\times L=I_{cm}\omega$

$I_{cm}=\frac{m(2L{)}^2}{12}=\frac{m{L}^2}{3}$

Kinetic Energy = $\frac{1}{2}m{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2=\frac{1}{2}(m\times (\frac{J}{m}{)}^2+I_{cm}\times \left(\frac{JL}{I_{cm}}{)}^2\right)$

$\Rightarrow KE=\frac{1}{2}(\frac{J^2}{m}+\frac{3J^2{L}^2}{m{L}^2})=2\frac{J^2}{m}$