Physics - Rotation Question with Solution | TestHub

The impulse $J$ causes both translational and rotational motion.

The kinetic energy is the sum of translational and rotational kinetic energy.

1. Calculate the change in linear momentum:

$J=\Delta p=M{v}_{CM}$

So, $v_{CM}=\frac{J}{M}$

2. Calculate the change in angular momentum:

The impulse is applied at one end, a distance $L/2$ from the center of mass.

$\tau \Delta t=J(L/2)=I\omega$

For a uniform rod about its center, the moment of inertia is $I=\frac{1}{12}M{L}^2$.

So, $J(L/2)=\frac{1}{12}M{L}^2\omega$

$\omega =\frac{J(L/2)}{\frac{1}{12}M{L}^2}=\frac{JL}{2}\cdot \frac{12}{M{L}^2}=\frac{6J}{ML}$

3. Calculate the total kinetic energy:

$KE=K{E}_{translational}+K{E}_{rotational}$

$KE=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}I{\omega }^2$

Substitute the expressions for $v_{CM}$ and $\omega$:

$KE=\frac{1}{2}M{\left(\frac{J}{M}\right)}^2+\frac{1}{2}\left(\frac{1}{12}M{L}^2\right){\left(\frac{6J}{ML}\right)}^2$

$KE=\frac{1}{2}M\frac{J^2}{M^2}+\frac{1}{24}M{L}^2\frac{36J^2}{M^2{L}^2}$

$KE=\frac{J^2}{2M}+\frac{36J^2}{24M}$

The final answer is $\frac{2J^2}{M}$.