Question

An 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 kinetic energy of the rod just after the impulse will be

• A. $\frac{J^2}{2m}$
• B. $\frac{J^2}{m}$
• C. $\frac{J^2}{m}$
• D. $\frac{6J^2}{m}$

Answer 1

Answer: (C)

Impulsive forces provides both linear momentum and angular momentum to the rod.
Let $\upsilon$ = linear velocity and
$\omega$ = angular velocity of rod after impulse is given.
Then, conservation of linear momentum gives,
$J=m{\upsilon }_{cm}$
$\Rightarrow {\upsilon }_{cm}=\frac{J}{m}$
Angular momentum is also provided by the impulse, so
$J\times L=I\omega$
$\Rightarrow JL=\frac{m{\left(2L\right)}^2}{12}\omega$
$\Rightarrow \omega =\frac{3J}{mL}$
Final kinetic energy of rod is sum of rotational kinetic energy and translational kinetic energy,
$\therefore$ Kinetic energy$\left(KE\right)$
$={\left(KE\right)}_{\text{translation}}+{\left(KE\right)}_{\text{rotation}}$
$=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}m\left(\frac{J^2}{m^2}\right)+\frac{1}{2}\frac{m{\left(2L\right)}^2}{12}\times {\left(\frac{3J}{mL}\right)}^2$
$=\frac{J^2}{2m}+\frac{36J^2}{24m}=\frac{4J^2}{2m}=\frac{2J^2}{m}$