JEE — Sphere Rolling UpHardQuestion 30362
Question
A solid sphere rolls up a rough incline (θ = 30°). The deceleration of its centre of mass is: (g = 10)
- A5 m/s²Correct
- B3.57 m/s²
- C7 m/s²
- D6 m/s²
Correct answer
5 m/s²
Explanation
For rolling up incline: a = g sinθ (1 + k) where for solid sphere k=2/5: a = 10×0.5×(1+2/5)... Actually a = g sinθ×(1+k) only in the context of deceleration. Correct formula: a = g sinθ/(1+k) for downward rolling. Going up, friction reverses: a = g sinθ(1+k)/(1+k) = g sinθ... Let me recalculate. For sphere rolling up, friction acts downhill (same direction as gravity component), so a = g sinθ + μg cosθ friction... Actually for pure rolling on incline: a = g sinθ/(1+I/mR²). For both up and down, |a| = g sinθ/(1+2/5) = 10×0.5/1.4 = 3.57 m/s².