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IRISSAK [1]
1 year ago
9

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 th

e 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
Physics
1 answer:
Stella [2.4K]1 year ago
5 0

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\times 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\times 2\times 2.5\ N\\\\F = 25  \ N

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.

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