An 80 kg solid sphere with a 0.1 m radius is suspended by a vertical wire attached to the ceiling of a room. If the sphere is given a twist and released, it to oscillates with a period of 5 s. How much torque is required to twist and hold the sphere at an angle of 1.5 radians

Torque = 0.7575 N.m

Explanation:

We are given;

Mass; m = 80kg

Radius = 0.1m

Period; T = 5s

Angle; θ = 1.5 radians

In simple harmonic motion, period T is given as;

T = 2π√(I/k)

Where,

I is rotational inertia of the object

k is torsion constant of the wire

Now, The rotational inertia has a formula;

I = (2/5)mr²

Where m is mass and r is radius.

Thus,

I = (2/5) x 80 x 0.1² = 0.32 kg.m²

Thus, we can now solve for k in the period equation ;

T = 2π√(I/k)

Make k the subject ;

k = 4π²I/(T²)

k = (4π² x 0.32)/(5²)

k = 0.505 N.m/rad

Now, the torque required to twist and hold is given by;

τ = kθ

Now, we know that k = 0.505 and θ = 1.5 rads

Thus, τ = 0.505 x 1.5 = 0.7575 N.m

(75°f − 32) × 5/9 = 23.889°c