Question

A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of 60° with the vertical, find the magnitude of the torque about the pivot.

Answer 1

Answer:

T=4.24 N.m

Explanation:

Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

F=0.5 kg* 9.8 m/$s^{2}$= 4.9 N

and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus

so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N

and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m

anothe way to do it is,

T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result