Sol. The kinetic energy of rotation of the flywheel is E = I ω 2 2 , where I = mR 2 , giving

ω 0 = =

= 1136 rad/s. Ans.

Measure time from the instant the flywheel is released, when it is rotating with angular velocity ω 0 . After its released the only horizontal force on the flywheel is the frictional force as shown in fig. The equations of motion are

I = – fR, m = f

At time t 1 when the flywheel stops slipping, let its angular velocity be ω 1 . The boundary conditions are ω = ω 0 , v = 0 at t = 0, ω = ω 1 , v = v 1 = R ω 1 at t = t 1 . The above equations integrate to give

I ( ω 1 – ω 0 ) = –fRt 1

mv = mR ω 1 = ft 1

Note that these equations can also be obtained directly by an impulse consideration. Solving these we have

ω 1 = , t 1 = ,

as I = mR 2 , f = µmg. The distance covered by the flywheel before it stops slipping is

S = t 1 2 = µg =

= 18290 m. Ans.

At t 1 the speed of the center of mass is

v 1 = R ω 1 = = 189.3 ms –1 . Ans.

At time 0 < t < t 1 , the equation of motion integrates to

I ( ω – ω 0 ) = –f Rt.

mv = ft.

At 0 < t < t 1 , the flywheel both slips and rolls. Only the slipping part of the motion casues dissipation of energy into heat. The slipping velocity is

v – R ω = – R ω 0

and the total dissipation of energy into heat is

Q = –

= + R ω 0 f t 1

= = 2.688 × 10 7 J.

This can also be obtained by considering the change in the kinetic energy of the flywheel:

Q = I ω 0 2 –

= . – = ,

Same as the above.