1.

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 NI 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

Answer»

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.



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