Question

A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s2, calculate the magnitude of F.

a. 35 N
a. 20 N
c. 25 N
d. 8 N

Answer 1

Given :

Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.

A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².

To Find :

The magnitude of F.

Solution :

Torque on hoop is given by :

`\tau =F* R\\\\I\alpha = FR\\\\MR^2\alpha = FR\\\\F = MR\alpha`( Moment of Inertia of hoop is MR² )

Putting value of M, R and α in above equation, we get :

`F=5* 2* 2.5\ N\\\\F = 25 \ N`

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.