Search any question & find its solution
Question:
Answered & Verified by Expert
A rod of length $l$ is held vertically stationary with its lower end located at a point $\mathrm{P}$, on the horizontal plane. When the rod is released to topple about $\mathrm{P}$, the velocity of the upper end of the rod with which it hits the ground is
Options:
Solution:
2616 Upvotes
Verified Answer
The correct answer is:
$\sqrt{3 \mathrm{~g} l}$
Here, potential energy of the metre stick will be converted into rotational kinetic energy.
Because centre of gravity of stick lies at the middle of the rod,
PE of metre stick $=\frac{m g l}{2}$
Rotational kinetic energy $E=\frac{1}{2} I \omega^{2}$
$I$ about point $A=\frac{m l^{2}}{3}$.
By law of conservation of energy
$$
m g\left(\frac{l}{2}\right)=\frac{1}{2} I \omega^{2}=\frac{1}{2} \frac{m l^{2}}{3}\left(\frac{v_{B}}{l}\right)^{2}
$$
By solving, we get $v_{B}=\sqrt{3 g l}$
Because centre of gravity of stick lies at the middle of the rod,
PE of metre stick $=\frac{m g l}{2}$
Rotational kinetic energy $E=\frac{1}{2} I \omega^{2}$
$I$ about point $A=\frac{m l^{2}}{3}$.
By law of conservation of energy
$$
m g\left(\frac{l}{2}\right)=\frac{1}{2} I \omega^{2}=\frac{1}{2} \frac{m l^{2}}{3}\left(\frac{v_{B}}{l}\right)^{2}
$$
By solving, we get $v_{B}=\sqrt{3 g l}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.