The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N
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 :
( Moment of Inertia of hoop is MR² )
Putting value of M, R and α in above equation, we get :

Therefore, the magnitude of force F is 25 N.
Hence, this is the required solution.
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