A weightless rod of length $l$ with a small load of mass m at the end is hinged at point $A$ as shown and occupies a strictly vertical position, touching a block of mass $M$. A light jerk sets the system in motion. For what mass ratio $\mathrm{M} / \mathrm{m}$ will the rod form an angle…

Question

A weightless rod of length $l$ with a small load of mass m at the end is hinged at point $A$ as shown and occupies a strictly vertical position, touching a block of mass $M$. A light jerk sets the system in motion. For what mass ratio $\mathrm{M} / \mathrm{m}$ will the rod form an angle $\theta=\pi / 6$ with the horizontal at the momentum of the separation from the block?

Aicharya · Accepted Answer

The loss in gravitional potential energy of the load $=$ Gain in kinetic energy of $\mathrm{b}$
$\begin{array}{l}
\Rightarrow \mathrm{mgl}\left(1-\sin 30^{0}\right)=\frac{1}{2} \mathrm{I} \mathrm{w}_{2}+\frac{1}{2} \mathrm{Mu}^{2} \\
\Rightarrow \frac{\mathrm{mgl}}{2}=\frac{1}{2} \mathrm{ml}^{2} \mathrm{w}^{2}+\frac{1}{2} \mathrm{Mu}^{2}
\end{array}$
Since the load and body have beenin contact upto this point, $1 \mathrm{w} \sin 30^{\circ}=\mathrm{u}$ or, $l \mathrm{w}=2 \mathrm{u}$
therefore the above equation becomes,
$\frac{\mathrm{mgl}}{2}=2 \mathrm{mu}^{2}+\frac{1}{2} \mathrm{Mu}^{2}$
Now consider the free body diagram of the load.
Centreipital acceleration is equal to the component of weight along the length of sin ce the body has just lost the contact.
$\begin{array}{l}
\Rightarrow \mathrm{mg} \sin 30^{\circ}=\mathrm{mw}^{2} \mathrm{r} \\
\Rightarrow \mathrm{w}^{2}=\frac{\mathrm{g}}{2 \mathrm{l}} \\
\Rightarrow \mathrm{u}^{2}=\frac{\mathrm{g} l}{8}=\frac{1}{2} \sqrt{\frac{\mathrm{gl}}{2}}
\end{array}$
$\mathrm{u} \sin \mathrm{g}$ this in above equation gives,
$\frac{\mathrm{M}}{\mathrm{m}}=4$

A weightless rod of length \(l\) with a small load of mass m at the end is hinged at point \(A\) as shown…

Solution