A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
The mass of the metre stick is 66 g.
Explanation
This problem applies the principle of rotational equilibrium from the chapter. When the metre stick balances at a new position after adding coins, the net torque about the new pivot point must be zero. The metre stick is uniform, so its centre of gravity lies at its geometric centre (50 cm mark). By taking moments about the new balance point (45 cm), we can equate the torques due to the metre stick's weight and the coins' weight to find the unknown mass.
Solution Steps
Step 1: Identify the given values
- Mass of two coins, m = 5 + 5 = 10 g
- Position of coins = 12.0 cm mark
- New balance point = 45.0 cm
- Centre of gravity of uniform metre stick = 50.0 cm
Step 2: Determine distances from the new pivot
- Distance of metre stick's CG from pivot = 50.0 - 45.0 = 5.0 cm
- Distance of coins from pivot = 45.0 - 12.0 = 33.0 cm
Step 3: Apply principle of moments for rotational equilibrium For equilibrium about the pivot at 45.0 cm: Anticlockwise moment = Clockwise moment m × 33.0 = M × 5.0 where M is the mass of the metre stick
Step 4: Calculate the mass of the metre stick M = (m × 33.0) / 5.0 M = () / 5.0 M = 330 / 5 M = 66 g