Question

A 9.9 kg stone is tied to a 6.2 m long string and swung around a circle at a constant angular velocity 4.9. Calculate the magnitude of the torque about the origin. * A uniform meter stick is placed on and parallel to the x axis, with the 50 cm mark at the origin. It is pivoted at the 50 cm mark. A -6.61 N force is applied at the 13 cm mark, and a -3.02 N force is applied at the 81 cm mark. Calculate the net torque on the rod. NOTES: Define up to be the +y-direction Remember to define the sign of the torque by the sense of rotation it causes (clockwise or counter-clockwise). Be careful of your units. T = 1.5095 IN m Your last answer was interpreted as follows: 1.5095

Answer 1

(1) Since there is no change in the angular velocity hence torque is constant

(2)Clockwise torque

   $\tau_{CW} = 3.02*(0.81-0.50) \\ = 0.9362~ N.m$

counterclockwise torque

                             $\tau_{CCW} = 6.61*(0.50-0.13) \\ = 2.4457~ N.m$

$\tau = 2.4457 - 0.9362 \\ \tau = 1.5095 ~N.m$in counterclockwise direction