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

(a) Refer to Fig.

Here, `M = 20kg, R = 20 cm = (1)/(5)m`
`F = 25N`
`alpha = ?, W = ?, K = ?`
Now, torque `tau = F xx R = 25 xx (1)/(5) = 5 Nm`
Moment of inertia of flywheel about its axis,
`I = (1)/(2)MR^(2) = (1)/(2) xx 20.0 ((1)/(5))^(2) = 0.4 kg m^(2)`
As `I alpha = tau`
`alpha = (tau)/(I) = (5)/(0.4) = 12.5 rad//s^(2)`
(b)Work done by the pull, unwinding `2` metre of the cord
`W = F xx x = 25 xx 2 = 50 J`
(c) As the wheel stars from rest, `omega_(1) = 0`. Angular displacement of wheel when `2m` string is unwound,
`theta = (l)/(R ) = (2)/(1//5) = 10 rad`.
If `omega_(2)` is final angular velocity of the wheel, then from
`omega_(2)^(2) + 2 alpha theta = 0 + 2(12.5) 10 = 250`
`K.E.` gained `= (1)/(2)I omega_(2)^(2) = (1)/(2) xx 0.4 xx 250`
`= 50J`
(d) The answers to (b) and (c) are the same, i.e., `K.E.` gained by the wheel = work done by the force. There is no loss of energy due to friction.