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A uniform, solid metal disk of mass 6.00 kg and diameter 29.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.28 N tangent to the rim of the disk to turn it by 3.32 ∘, thus twisting the wire. You now remove this force and release the disk from rest.

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jolis1796

Answer:

k = 21.42 Nm / rad

Explanation:

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

τ = - k*θ

We know that when we apply the force 4.28 N the system is in equilibrium with a twist angle

θ = 3.32° = 0.0579 rad.

The torque in this case is τ = R*F = 0.29m*4.28N = 1.2412 Nm.

Then

k =  τ / θ

⇒ k = (1.2412 Nm) / (0.0579 rad) = 21.42 Nm / rad

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