Question

A small ball of mass 0.75 kg is attached to one end of a 1.25-m-long massless rod, SSM SSM ILW and the other end of the rod is hung from a pivot.When the resulting pendulum is 30# from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

Answer 1

The "magnitude of the torque" is "4.59375 Nm".

Explanation:

The "gravitational force" on the "bob of the pendulum" is m × g

m = 0.75 kg

$\text { (Assuming "acceleration due to gravity" to be } \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { ) }$

m = 0.75 × 9.8 = 7.35 N  

The "torque of this force" around "the pivot" is given by the "vector product" of the "radial displacement vector" with "the force vector".

The "magnitude of this torque" is given by   "r × F × sin A"  

Where, r = the "radial distance" = 1.25 m  

F = "gravitational force" = 7.35 N

$A=\text { is the angle between them }=30^{\circ}$

Thus the "magnitude of the torque" is = 1.25 × 7.35 × sin30

The "magnitude of the torque" = 9.1875 × 0.5  

The "magnitude of the torque" = 4.5375

Therefore, the "magnitude of the torque" is 4.59375 Nm