Question

A boy balances a pair of equal masses on a uniform massless rod of length 1 m at its midpoint. 

1. Prove that the centre of gravity lies exactly at the midpoint. 

2. Are there any variations if masses are changed in equal proportion? 

3. Calculate the balancing point if only one of the mass is doubled.

Answer 1

consider the free body diagram (FBD) shown in the figure. N is the normal force applied by boy. Since the system is in rotational equilibrium about 0.

∴ mgl = mg (L – l)

∴ 2 l = L ⇒ l = \(\frac{L}{2}\)

Hence the centre of gravity lies exactly at the mid pointmid point of rod.

2. No, the centre of gravity does not change. 

3. Consider the FBD

The rod is balanced about 0 (balancing point)

∴ 2mgl = mg (L – l)

∴ 3L = L

⇒ l = \(\frac {L}{3}\)from 2m.