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

Given M=20 kg
R=20cm=0.2cm
F=25N
(a) Torque, `tau=F.R=25xx0.2=5.0Nm`
Moment of inertia of the wheel about its axis is
`I=(MR^(2))/(2)=(20xx(0.2)^(2))/(2)=0.4kgm^(2)`
We know, `tau=Ialpha`
`:.alpha=(tau)/(I)=(5.0)/(0.4)=12.5rad//s^(2)`
(b) Work done by the pull, W=F.s
`=25xx2=50s`
(c) Angular displacement of the wheel,
`theta=("Length of string unwounded")/("Radius of wheel")=(2)/(0.2)=10` radians
Initial angular veocity of wheel, `omega_(0)=0`
Using `omega^(2)=omega_(0)^(2)+2alpha,theta` we get
`omega^(2)=0+2xx12.5xx10=250(rad//s)^(2)`
`:.` Kinetic energy gained, `K=(1)/(2)Iomega^(2)`
`=(1)/(2)xx0.4xx250=50J`
(d) Both the answers are same, i.e. kinetic energy gained = work done by the force