A ring is at the top of the incline of angle and height ‘h’. If it is left from rest and goes down rolling purely, find its speed at the bottom.
(a) √ (2gh) m/s
(b) √ (gh/sinθ) m/s
(c) √ (ghsinθ) m/s
(d) √ (gh) m/s
I got this question during an interview.
My question is based upon Rolling Motion in portion System of Particles and Rotational Motion of Physics – Class 11
Select the correct answer from above options
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The correct option is (d) √ (gh) m/s
Explanation: Friction will be acting on the ring, which is the reason it is able to roll purely. But this friction will not do any work on the ring. The potential energy at the top will be converted into kinetic energy at the bottom. Kinetic energy is given by: 1/2MV^2 + 1/2Iw^2. I for a ring = MR^2 & w = V/R.
∴ kinetic energy = 1/2MV^2 + 1/2MV^2 = MV^2.
Potential energy at the top = Mgh.
Therefore, Mgh = MV^2 OR V = √(gh) m/s.