Question

A solid uniform cylinder with a radius of 0.3m and a mass of 12kg is free to rotate about it axis, which is supported by frictionaless bearing a constant tangential force of 10N is applied to the cylinder calculate (i)the torque apply to the cylinder (ii) what's the angular acceleration of the cylinder (iii) what's the angular velocity of the cylinder 3 secs after the force applied.

Answer 1

Solution:

(i)the torque apply to the cylinder:

τ = FR = 10(0.3) = 3 Nm

(ii) what's the angular acceleration of the cylinder:

α = $\dfrac{t}{l}$ = $\dfrac{3}{ \tfrac{1}{2} \cdot 12 \cdot 0.3^{2} } = 5.56 ( \tfrac{rad}{sec^{2}} )$

(iii) what's the angular velocity of the cylinder 3 secs after the force applied:

ω = αt = 5.5555(3) = 16.6666... ≈ 17 $( \tfrac{rad}{sec} )$