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} $