Question

The center of mass of a 0.2 kg (non-uniform) meter stick is located at its 46-cm mark. What is the magnitude of the torque (in Nm) due to gravity if it is supported at the 26-cm mark? (Use g = 9.79 m/s2).

Answer 1

weight would act at the centre of mass.

Hence torque due to this force would be mg * (distance between point of suspension and centre of mass) = (0.2)(9.79)(0.2) = 0.39 Nm