A solid uniform sphere (moment of inertia 2/5 M R^2 about its centre)…
Question
A solid uniform sphere (moment of inertia 2/5 M R^2 about its centre) is released from rest and rolls without slipping down a rough incline, descending a vertical height of 7 m. At the bottom it smoothly enters a frictionless circular loop track. Because the loop is frictionless, the sphere's spin rate cannot change there. Find the maximum vertical height (in metres) the sphere rises above the bottom while on the frictionless loop.
Step-by-step solution
Rolling down the rough incline, energy conservation: M g h = (1/2) M v^2 + (1/2) I omega^2 with I = (2/5)M R^2 and v = omega R, so M g h = (1/2)M v^2 (1 + 2/5) = (7/10) M v^2, giving v^2 = 10 g h / 7.
On the frictionless loop there is no friction torque, so the spin (rotational KE) stays constant; only the translational KE (1/2)M v^2 converts to potential energy: (1/2)M v^2 = M g h'.
Final answer5
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Physics Mechanics solutions