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Grade: 11
        
A 95.2-kg solid sphere with a 14.8-cm radius is suspended by a vertical wire attached to the ceiling of a room. A torque of 0.192 N· m is required to swist the sphere through an angle of 0.850 rad. Find the period of oscillation when the sphere is released from this position.
3 years ago

Answers : (2)

Jitender Pal
askIITians Faculty
365 Points
							For small twists the restoring torque τ is proportional to the angular displacement θ, so that
τ = κ θ
So from the above equation τ = κ θ, the torsional constant κ will be,
κ = τ/ θ
To find out the torsional constant κ, substitute 0.192 N.m for τ and 0.850 rad for θ in the equation κ = τ/ θ,
κ = τ/ θ
= 0.192 N.m/0.850 rad
= 0.226 N.m
We know that the rotational inertia I of the solid sphere is,
I = 2/5 MR2
Here, M is the mass of the body and R is the radius.
To obtain the rotational inertia I of the solid sphere, substitute 95.2 kg for M and 14.8 cm
235-1738_4.PNG
3 years ago
pa1
357 Points
							
235-1738_4.PNGHere, M is the mass of the body and R is the radius.To obtain the rotational inertia I of the solid sphere, substitute 95.2 kg for M and 14.8 cm2For small twists the restoring torque t is proportional to the angular displacement ?, so thatt = ? ?So from the above equation t = ? ?, the torsional constant ? will be,? = t/ ?To find out the torsional constant ?, substitute 0.192 N.m for t and 0.850 rad for ? in the equation ? = t/ ?,? = t/ ? = 0.192 N.m/0.850 rad = 0.226 N.mWe know that the rotational inertia I of the solid sphere is,I = 2/5 MR
3 years ago
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