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JEE Main 2014 · 9 Apr · Shift 1, Physics question 6

A cylinder of mass McM_c and sphere of mass MsM_s are placed at points A and B of two inclines, respectively. (See Figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio sin⁡θcsin⁡θs\dfrac{\sin\theta_c}{\sin\theta_s} is :

A vertical wall AC with two inclines meeting the ground at D: AD (steeper, angle θ_c at D) carries the cylinder M_c starting at A, and BD (shallower, angle θ_s, B lower on the wall) carries the sphere M_s.
Choose one option
Answer and worked solution

Official answer: D

Rolling down an incline: a=gsin⁡θ1+k2/r2a=\dfrac{g\sin\theta}{1+k^2/r^2}.

  • Cylinder (k2/r2=12k^2/r^2=\frac12): a=23gsin⁡θca=\frac23g\sin\theta_c.
  • Sphere (25\frac25): a=57gsin⁡θsa=\frac57g\sin\theta_s.

Equal accelerations give
sin⁡θcsin⁡θs=5/72/3=1514.\frac{\sin\theta_c}{\sin\theta_s}=\frac{5/7}{2/3}=\frac{15}{14}.

Topics:Rolling motionMoment of inertia
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