
From energy conservation
mgh=21mv02+21Iω2
⇒mgh=21mv02+21×52ma2×a2v02
⇒gh=21v02+51v02
⇒gh=107v02
From triangle, sinθ=lh
Then h=lsinθ
⇒lsinθ=107gv02
Thus, the sphere will travel distance, l=107gsinθv02.
A sphere of radius a and mass m rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Held on 25 Feb 2021 · Verified 6 Jul 2026.
52gsinθv02
10gsinθ7v02
5gsinθv02
2gsinθv02
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