Answer in Mechanics | Relativity for Ojugbele Daniel #138998
March 12th, 2023
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}t} = 5.56 ( tfrac{rad}{sec^{2}}t)”
(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}t)”
