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 O. If a horizontal impulse P is imparted to the rod at a distance x = L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of n is ______.

Answer 1

Correct answer is : 18

Linear impulse \(\int F d t=\Delta\) momentum

\(=\mathrm{m}\left(\mathrm{V}_{\mathrm{cm}}-0\right) \)

\( \mathrm{P}=\mathrm{m}\left(\omega \mathrm{r}_{\mathrm{cm}}\right)\)

\(=\mathrm{m} \omega\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right) \)

\(\mathrm{P}=\mathrm{m} \omega\left(\frac{3 \mathrm{~L}}{2}\right)\) .....(i)

Angular impulse \(\int \, \tau dt = \Delta \) angular momentum

\(\int \mathrm{r} \times \mathrm{Fdt}=\Delta \mathrm{L}\)

\(\mathrm{r} \times \int \mathrm{Fdt}=\mathrm{I}(\omega-0) \text {, and } \mathrm{I}\) is moment of inertia about axis of rotation.

\( \left(\mathrm{L}+\frac{\mathrm{L}}{2}+\mathrm{x}\right) \times \mathrm{P}=\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2\right) \omega\)

\(=\left(\frac{\mathrm{mL}^2}{12}+\mathrm{m}\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right)^2\right) \omega\)

\( \left(\frac{3 \mathrm{~L}}{2}+x\right) P=\mathrm{mL}^2\left(\frac{1}{12}+\left(\frac{3}{2}\right)^2\right) \omega\)

\(\left(\frac{3 L}{2}+x\right) P=\mathrm{mL}^2\left(\frac{7}{3}\right) \omega \) .....(ii)

Divide eq.- (i) & (ii)

\(\left(\frac{3 \mathrm{~L}}{2}+x\right)=\frac{\mathrm{L}\left(\frac{7}{3}\right)}{\left(\frac{3}{2}\right)} \)

\(\frac{3 \mathrm{~L}}{2}+x=\mathrm{L}\left(\frac{14}{9}\right)\)

\(x=\frac{L}{18}\)