Question

A rod of mass m & length L is hinged at it's upper end. It can rotate in vertical plane. It is given angular
velocity omega so that it can complete vertical circle. Find (a)omega(angular velocity) (b) Tension at centre of rod at initial moment.

Answer 1

Answer:

(a) $\bold{\omega=\sqrt{\frac{3 g}{l}}}$

(b) $\bold{T=mg}$

Explanation:

(a) Potential energy of the rod:

$\Delta P E=\frac{m g l}{2}$

Kinetic energy of the rod:

$\Delta K E=\frac{1}{2} I \omega^{2}$

Where,

I = Moment of Inertia

ω = Angular velocity

The moment of inertia of a rod fixed in the upper end is:

$I=\frac{m l^{2}}{3}$

Now, the kinetic energy becomes,

$\Delta K E=\frac{1}{2}\left(\frac{m l^{2}}{3}\right) \omega^{2}$

The change in potential energy is equal to change in kinetic energy when one circle is completed.

$\Rightarrow \Delta P E=\Delta K E$

$\Rightarrow \frac{m g l}{2}=\frac{1}{2}\left(\frac{m l^{2}}{3}\right) \omega^{2}$

$\Rightarrow \omega^{2}=\frac{3 g}{l}$

$\therefore \omega=\sqrt{\frac{3 g}{l}}$

(b) At the center of the rod, the tension at the initial moment is weight of the rod before rotation.

$T=m g$

Answer 2

Answer:√(3g/l)

Hint change in potential energy is always equal to change in kinetic energy