31. A cord of negligible mass is wound round the rim of a fly wheel of mass 24 kg and radius 24 cm. A steady pull of 30 N is applied on the cord. 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 2m of the cord is unwound.
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).
Answers
Explanation:
Given : mass of flywheel M = 24 kg
radius r = 24 cm = 0.24 m
Force on chord F = 30 N
a) M.O.I of fly wheel about its axis of rotation is
I = Mr²/2 = 24 x( o.24)²/2 N m²
= 24 x 0.24 x 0.12 N m²
Torque on the flywheel τ = F x r
= 30 x 0.24
τ = I α
Angular accln α = τ/I
=( 30 x 0.24 )/(24 x 0.24 x 0.12)
= 10.42 rad/s²
b) W.D by the pull = F x s
= 30 x 2
= 60 N m
c) loss in PE during unwounding = W.D by pull
According to law of conservation of energy
Loss in PE = 1/2 M v² + 1/2 I ω²
60 = 1/2 M v² + 1/2 [Mr²/2]{v²/r²]
60 = 1/2 Mv² + 1/4 Mv²
= 3/4 Mv²
∴ Mv²/4 = 60/3 = 20 J
∴ K.E of fly wheel = 1/2 Iω² = 1/4 Mv² = 20J