A cord of negligible mass is wound round the rim of a flywheel of mass `20 kg` and radius `20 cm`. A steady pull of `25 N` is applied on the cord as shown in Fig. The flywheel is mounted on a horizontal axle with frictionless bearings.

(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when `2 m` of the cord is unwound.
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel stars from rest.
(d) Compare answers to parts (b)and (c).

Explanation

Moment of inertia of the flywheel about its axis, i.e.,
`I=1/2MR^(2) =1/2(20 kg) (0.2 m^(2)) = 0.4 kg m^(2)`
(i) Angular acceleration,
`alpha =tau/I = (FR)/I =((25N)(0.20m))/(0.4 kg m^(2)) = 12.5 rad//s^(2)`
(ii) Work done by the pull in unwinding 2m of the cord, i,e.,
W= `(25 N) (2m) = 50 J`
(iii) As `omegao^(2) = omega_(0)^(2) + 2alpha theta`,
`omega^(2) = 2alpha theta =2(12.5 rad//s^(2))((2m)/(0.2m) rad) = 250 (rad//s^(2))`
(since, `omega_(0)=0` and `theta = 1//r`)
Kinetic energy gained by the wheel.
`K=1/2 Iomega^(2) =1/2(0.4 kg m^(2))(250 (rad //s^(2))) = 50 J`
(iv) W=K, i.e., there is no loss of energy due to friction.