Question #208337

A flywheel that is a solid disc and initially at rest, has a force of 20.0 N applied tangentially to its rim for 10.0 s. The flywheel is 365 mm in diameter and has a mass of 30.5 kg, and the bearings have a friction torque of 0.450 N m.

a) Determine the torque applied to the flywheel

b) Calculate the moment of inertia of the flywheel

c) What is the angular acceleration of the flywheel during the 10 s?

d) How much kinetic energy does the flywheel have at t=10 s?


Expert's answer

(a) The applied torque:


τa=Fd2=200.3652=3.65 Nm.\tau_a=F\frac d2=20·\frac{0.365}{2}=3.65\text{ N}·\text m.

(b) The moment of inertia:


I=18md2=0.508 kgm2.I=\frac{1}{8}md^2=0.508\text{ kg}·\text{m}^2.

(c) The angular acceleration is


α=τnetI=τaτfI, α=3.650.450.508=6.30 rad/s2.\alpha=\frac{\tau_\text{net}}{I}=\frac{\tau_a-\tau_f}{I},\\\space\\ \alpha=\frac{3.65-0.45}{0.508}=6.30\text{ rad/s}^2.

(d) Find the angular velocity at t=10 s:


ω=αt=63 rad/s.\omega=\alpha t=63\text{ rad/s}.

The kinetic energy:


E=12Iω2=1010 J.E=\frac 12 I\omega^2=1010\text{ J}.


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