A ring of diameter 1m is rotating about a central axis perpendicular to its diameter. It is rotating with a speed of 10rad/s. What force should be applied on it tangentially to stop it in exactly 3 rotations? Let the mass of the ring be 2kg.
(a) 10 N
(b) 25/3π N
(c) 50/3π N
(d) 12.5/3 N
I had been asked this question by my school principal while I was bunking the class.
Origin of the question is Kinematics of Rotational Motion about a Fixed Axis in portion System of Particles and Rotational Motion of Physics – Class 11
Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
Right option is (b) 25/3π N
The best I can explain: Let w be the angular velocity & a be the angular acceleration.
Moment of inertia of the ring = MR^2
= 2*0.25 = 0.5kgm^2.
Angular distance = 3 rotations = 3*2π rad = 6π rad.
Using, w^2 = w0^2 + 2aθ, we get:
0 = 100 – 2a*6π.
a = 100/12π.
Torque = Ia = rF
∴ F = Ia/r = 0.5*100/(12π*0.5) = 25/3π N.