Question 407: Centre of Mass and Rotational Motion

Question

A uniform rod $A B$ of mass $2 \mathrm{~kg}$ and length $30 \mathrm{~cm}$ at rest on a smooth horizontal surface. An impulse of force $0.2 \mathrm{~Ns}$ is applied to end B. The time taken by the rod to turn through at right angles will be $\frac{\pi}{x} \mathrm{~s}$, where $x=$ _______ .

Accepted Answer

Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$
$$
\mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s}
$$
Angular impuls $(\vec{M})$
$$
\begin{aligned}
& \vec{M}_c=\int 	au d t \\
& =\int F \frac{L}{2} d t \\
& =\frac{L}{2} \int F d t=\frac{L}{2} 	imes J \\
& =\frac{0.3}{2} 	imes 0.2 \\
& =0.03
\end{aligned}
$$
$\begin{aligned} & I_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}=\frac{2 	imes(0.3)^2}{12}=\frac{0.09}{6} \\ & \mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}ight) \\ & 0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}ight) \\ & \omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}\end{aligned}$
$	heta=\omega \mathrm{t}$
$\mathrm{t}=\frac{	heta}{\omega}=\frac{\pi}{2 	imes 2}=\frac{\pi}{4} \mathrm{sec}$
$X=4$

Centre of Mass and Rotational Motion

A uniform rod A B of mass 2 \mathrm{~kg} and length 30 \mathrm{~cm} at rest on a smooth horizontal surface. An impulse of force 0.2 \mathrm{~Ns} is applied to end B. The time taken by the rod to turn through at right angles will be \frac{\pi}{x} \mathrm{~s}, where x= _______ .

Choose the correct answer: