Question #118177

A flywheel has a moment of inertia of 1.6 Ã— 10âˆ’3 kg.m2. When a constant torque is applied, it reaches an angular velocity of 1200 rev/min in 15 s. Assuming it started from rest, find:

i) the angular acceleration

ii) the unbalanced torque applied

iii) the angle turned through 15 s

iv) the work done on the flywheel by the torque

i) the angular acceleration

ii) the unbalanced torque applied

iii) the angle turned through 15 s

iv) the work done on the flywheel by the torque

Expert's answer

Given, Moment of inertia "I=1.6 \u00d7 10^{\u22123} kg.m^2" ,

Initial angular speed "\\omega_0=0" ,final angular speed "\\omega=1200\\times \\frac{2\\pi}{60}=40\\pi \\:rad\/s" ,time "t=15s" ,

i). Angular acceleration

"\\alpha=\\frac{\\omega-\\omega_0}{t}\\implies \\alpha=\\frac{40\\pi}{15}=\\frac{8\\pi}{3} \\: rad\/s^2"ii).

Unbalanced torque

"\\tau=I\\alpha\\\\\n\\implies \\tau=1.6\\times 10^{-3}\\times \\frac{8\\pi}{3}=13.40\\times 10^{-3}\\:Nm"iii). Let the required angle is "\\theta" , thus

"\\omega^2=\\omega_0^2+2\\alpha \\theta\\\\\n\\implies \\theta=\\frac{\\omega^2-\\omega_0^2}{2\\alpha}\\\\\n\\implies \\theta=\\frac{(40\\pi)^2}{2\\frac{8\\pi}{3}}=942.47 \\: rad"Alternatively, we can apply

"\\theta=\\omega_0t+\\frac{1}{2}\\alpha t^2"

and eventually we will get the same result.

iv).

By applying work energy theorem for rotational motion we get,

Thus we get,

"W_{rot}=\\frac{1}{2}\\times 1.6\\times 10^{-3} (40\\pi)^2=12.63J"Hence we done.

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