Question

Consider a 0. 5-kg meter stick with a 1. 0 kg mass attached at the 50 cm mark and a 3. 0 kg mass at the very end. Where on the meter stick is the center of mass located?.

Answer 1

The effective mass at the 50 cm mark will be 1.5 kg because the meter stick is uniform and .5 kg can be considered to be acting at 50 cm

3 * (50 - x) = 1.5 * x

Balancing torques where x is the distance from the 50 cm mark

150 - 3 x = 1.5 x

x = 150 / 4.5 = 33.3

x would be at 50 - 33.3 = 16.7

Check: 3 * 16.7 = 1.5 * 33.3

50.1 = 50 within rounding