Question

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

Answer 1

Hi there!

Recall the equation for Torque:
$\large\boxed{\Sigma \tau = rFsin\phi}$

In words, the torque is the force multiplied by the distance from the LINE OF ACTION to the PIVOT POINT.

In this instance:

F = Mg = 0.75 · 9.8 = 7.35 N

r = 1.25 m

sin(30) = 0.5

Plug in the values:

$\Sigma \tau = (1.25)(7.35)(0.5) = \boxed{4.59 Nm}$