Question

A rod of length $1$ meter is standing vertically, when its other end touches the ground without slipping then the speed of other end will be:-

Answer 1

The center of gravity of rod lies at 1/2m i.e. at the center of the rod.

Thus, when the rod was vertical the total potential energy in it was $mgh=4.9m$

When it just touched the ground its entire potential energy had been converted into kinetic energy.

The kinetic energy was there involved in rotating the rod.

Let the rod touch the floor with an angular velocity $\omega$. At time of touching the ground entire energy is kinetic.

By energy conservation

Initial potential energy = Final kinetic energy

$4.9m=\frac{1}{2}I{\omega }^2......\left(1\right)$

The moment of inertia of a rod perpendicular to its axis, along its center of gravity

$I=\frac{m{L}^2}{12}$

From (1)

$4.9m=\frac{1\times m\times {\omega }^2}{2\times 12}$

$117.6={\omega }^2$

$\omega =10.95$

Linear velocity at the end of the rod will be

$v=\omega \times r$

$=10.95\times \frac{1}{2}$

$=5.42m/s$