A uniform, solid metal disk of mass 6.0 kg and diameter ...

A uniform, solid metal disk of mass 6.0 kg and diameter 2.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4 N tangent to the rim of the disk to turn it by 5 degrees, thus twisting the wire. You now remove this force and release the disk from rest.

a.  What is the torsion constant for the metal wire? 

b.  Write the equation of motion for twist angle of the disk.

Solution

a. The torque is related to the torsion constant (k) by the equation: 

τ = -kθ

We know that when we apply the force 4 N the system is in equilibrium with a twist angle: 

θ = 5^o = 0,087 rad 

The torque in this case is:

τ = FR = 4 * 0,01 = 0,04 Nm 

Then:

k = τ / θ = 0,04 / 0,087 = 0,46 Nm / rad

b. To write equation of motion we need to find the moment of inertia of the disk. It has the following expression:

I = ½ mR² = ½ *6 * 0,01² = 0,00024 kg * m²

Then:

I θ = -k θ

0,00024 * θ = - 0,46 * θ

θ = - 1917 * θ

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