Question 741: Centre of Mass and Rotational Motion

Question

A thin uniform rod of length $L$ and certain mass is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $\mathrm{O}$. If a horizontal impulse $P$ is imparted to the rod at a distance $x={L \over n}$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $\mathrm{O}$, with the rod remaining aligned with the string. In such a case, the value of $n$ is ___________.

Accepted Answer

Linear impulse $$\int \mathrm{Fdt}=\Delta$$ momentum
$$\begin{aligned}
& =\mathrm{m}\left(\mathrm{V}_{\mathrm{cm}}-0ight) \
& \mathrm{P}=\mathrm{m}\left(\omega \mathrm{r}_{\mathrm{cm}}ight) \
& =\mathrm{m} \omega\left(\mathrm{L}+\frac{\mathrm{L}}{2}ight) \
& \mathrm{P}=\mathrm{m} \omega\left(\frac{3 \mathrm{~L}}{2}ight) \quad 	ext{... (i)}
\end{aligned}$$
Angular impulse $$\int 	au \mathrm{dt}=\Delta$$ angular momentum
$$\int \mathrm{r} 	imes \mathrm{Fdt}=\Delta \mathrm{L}$$
$$\mathrm{r} 	imes \int \mathrm{Fdt}=\mathrm{I}(\omega-0)$$, and I is moment of inertia about axis of rotation.
$$\begin{aligned}
\left(\mathrm{L}+\frac{\mathrm{L}}{2}ight. & +\mathrm{x}) 	imes \mathrm{P}=\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2ight) \omega \
& =\left(\frac{\mathrm{mL}^2}{12}+\mathrm{m}\left(\mathrm{L}+\frac{\mathrm{L}}{2}ight)^2ight) \omega
\end{aligned}$$
$$\begin{aligned}
\left(\frac{3 L}{2}+xight) P & =m L^2\left(\frac{1}{12}+\left(\frac{3}{2}ight)^2ight) \omega \
\left(\frac{3 L}{2}+xight) P & =m L^2\left(\frac{7}{3}ight) \omega \quad 	ext{... (ii)}
\end{aligned}$$
Divide eq.-(i) & (ii)
$$\begin{aligned}
& \left(\frac{3 L}{2}+xight)=\frac{L\left(\frac{7}{3}ight)}{\left(\frac{3}{2}ight)} \
& \frac{3 L}{2}+x=L\left(\frac{14}{9}ight) \
& x=\frac{L}{18}
\end{aligned}$$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type