Question

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.

Answer 1

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)