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A circular disc reaches from top to bottom of an inclined plane of length l. When it slips down the plane, it takes t s. When it rolls down the plane then it takes ( 2 )1/2t s, where is .
3
Both start from rest and cover the same length l: l = 12at2, so t 1 a.
Slipping (no friction): a = g
Rolling disc ( k2R2 = 12): a' = g 1 + k2R2 = 23g
t' = t aa' = ( 32 )1/2t
Comparing with ( 2 )1/2t gives = 3.
Answer: 3
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