Question

A uniform disc of mass 2 kg and radius 1 m is mounted on an axle supported on fixed frictionless bearings. A light chord is wrapped around the rim of the disc and mass of 1 kg is tied to the free end. If it is released from rest-

• A. the tension in the chord is 5 N
• B. in first four seconds the angular displacement of the disc is 40 rad
• C. the work done by the torque on the disc in first four seconds is 200 J
• D. the increase in the kinetic energy of the disc in the first four seconds is 200 J

Answer 1

Answer: (A), (B), (C), (D)

the increase in the kinetic energy of the disc in the first four seconds is 200 J

By FBD of particle


mg – T = ma
10–T = a ...(i)
By FBD of disc
$TR=I\alpha =I\frac{a}{R}\Rightarrow T=\frac{M{R}^2}{2}\frac{a}{R^2}$
$T=\frac{Ma}{2}=a$
By eq.(i) and (ii)
(A) a = 5 $\frac{m}{s^2}$ and T = 5 N and
$\alpha =\frac{a}{R}=5\frac{rad}{s^2}$
(B) For angular displacement of disc:
$\theta =\omega t+\frac{1}{2}\alpha t^2$
$\theta =\frac{1}{2}\times \alpha t^2=\frac{1}{2}\times 5\times (4{)}^2$
$\theta =\frac{1}{2}\times 5\times 16=40rad$
(C) Work done by torque
$=\int \tau d\theta =\tau \int d\theta =5\times 40=200J$
(A) $\Delta K.E=\Delta W=200J$
$K_2-K_1=200J$