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JEE Advanced 2022 Paper 1 · Physics · Rotational Motion · medium

A uniform solid sphere of mass m and radius R is released from rest on a rough incline with tan θ=0.7 and rolls down without slipping.

One or more options correct

A uniform solid sphere of mass mm and radius RR is released from rest on a rough incline with tan⁡θ=0.7\tan\theta=0.7 and rolls down without slipping. Choose the correct option(s).
  1. A. Its linear acceleration is 57 gsin⁡θ\tfrac57\,g\sin\theta.
  2. B. The minimum coefficient of static friction needed is 0.20.2.
  3. C. After its centre has descended a vertical height hh, its speed is 10gh/7\sqrt{10gh/7}.
  4. D. Over this descent the work done by static friction on the sphere is −27mgh-\tfrac27mgh.
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Answer: A, B, C

Rolling gives a=gsin⁡θ1+I/mR2=57gsin⁡θa=\dfrac{g\sin\theta}{1+I/mR^2}=\tfrac57g\sin\theta, so (A) is true. Friction f=27mgsin⁡θ≤μmgcos⁡θf=\tfrac27mg\sin\theta\le\mu mg\cos\theta gives μ≥27tan⁡θ=0.2\mu\ge\tfrac27\tan\theta=0.2, so (B) is true. Energy: mgh=12mv2(1+25)mgh=\tfrac12mv^2\left(1+\tfrac25\right), so v=10gh/7v=\sqrt{10gh/7}, (C) is true. (D) is false: the point of contact is instantaneously at rest, so static friction does zero net work; its negative translational work is cancelled by its positive rotational work.
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