Question

A uniform thin rod of length $L$, mass $m$ is lying on a smooth horizontal table. A horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

• A. $\frac{P^2}{M}$
• B. $\frac{7P^2}{8M}$
• C. $\frac{13P^2}{2M}$
• D. $\frac{2P^2}{M}$

Answer 1

Answer: (D)

$P=(mv-0)$
angular impulse $J=P\times \frac{L}{2}=L_f-0$
$L_f=\frac{PL}{2}$
$KE=K_r+K_t=\frac{P^2}{2m}+\frac{L_f^2}{2I}$
$=\frac{P^2}{2m}+\frac{{\left(P\frac{L}{2}\right)}^2}{2\times \frac{m{L}^2}{12}}$
$=\frac{P^2}{2m}+\frac{P^2{L}^2}{4\times \frac{m{L}^2}{6}}$
$=\frac{P}{2m}+\frac{3P^2}{2m}=\frac{2P^2}{m}$