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Physics Question on System of Particles & Rotational Motion

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg3 \,kg and radius 40 cm40 \,cm. If the rope is pulled with a force of 30 N30 \,N, then the angular acceleration produced in the cylinder is

A

15 rad s−215\,rad\, s^{-2}

B

20 rad s−220\,rad\, s^{-2}

C

25 rad s−225\,rad\, s^{-2}

D

30 rad s−230\,rad\, s^{-2}

Answer

25 rad s−225\,rad\, s^{-2}

Explanation

Solution

Here, M=3 kgM = 3 \,kg
R=40 cm=40×10−2 mR = 40 \,cm = 40 \times 10^{-2}\,m
Force applied, F=30 NF = 30 \,N
Torque, τ=FR=(30N)(40×10−2 m)=12 N m\tau = FR = (30 N) (40 \times 10^{-2}\, m) = 12\, N \,m
Moment of inertia of hollow cylinder about its axis is
I=MR2=(3 kg)(40×10−2 m)2=0.48 kg m2I = MR^2 = (3\, kg) (40 \times 10^{-2}\, m)^2 = 0.48 \,kg\, m^2
Let a is the angular acceleration produced.
As τ=Iα\tau = I\alpha
∴α=τI\therefore\alpha = \frac{\tau}{I}
=12 N m0.48 kg m2=\frac{ 12 \,N \,m}{0.48 \,kg\, m^{2}}
=25 rad s−2= 25\, rad\, s^{-2}