A mass of 200 gm is attached to a cord wound over a flywheel of mass 10 kg and
circumference 75 cm with an axle diameter 3 cm. A mark is made at a distance 2 cm in
radial direction from the edge of the flywheel. Calculate the angular velocity and linear
velocity of this mark when the time is 20 s after the mass is released from the rest.
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Answer:
Angular velocity is 10.2(s−1)
linear velocity is 7.4(m/s)v
Explanation:
Energy is stored mechanically in a flywheel as kinetic energy.
Kinetic energy in a flywheel can be expressed as
Ef = 1/2 I ω2
where
Ef = flywheel kinetic energy (Nm, Joule, ft lb)
I = moment of inertia (kg m2, lb ft2)
ω = angular velocity (rad/s)
Formula for flywheel angular acceleration is;
Flywheel angular acceleration ϵ=Rmg/(J+mR2)
Putting values;
=0.51(s−2)
which means the angular velocity is ω=ϵt=0.51⋅20=10.2(s−1)
and the linear velocity is v=ω(R−d)=10.2⋅0.73=7.4(m/s)
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