A body of mass m, radius R, and radius of gyration K slides down a smooth inclined plane of inclination θ from height h and reach the bottom with speed v. Now the same inclined plane is made rough and same body rolls down the inclined plane without slipping and reach the ground with speed V. Find the ration v: V.

Solution

Acceleration of body when it slides down the inclined plane without rolling is,

a₁ =gsinθ

The speed of the body when it reaches the ground by slipping is,
$straight v equals square root of 2 straight a subscript 1 straight I end root equals square root of 2 gIsin space straight theta end root$
The acceleration of body when it rolls down the inclined plane without slipping is,

$straight alpha subscript 2 space equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style 1 over mR squared end style end fraction space space space space space space equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style mK squared over mR squared end style end fraction space space space space space space equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight K squared over straight R squared end style end fraction$
The speed of the body when it reaches the ground by rolling is,
$straight V equals square root of 2 straight a subscript 1 straight I end root equals square root of fraction numerator 2 gI space sinθ over denominator 1 plus begin display style straight K squared over straight R squared end style end fraction end root space Now comma space the space ratio space is space given space by comma straight v over straight V equals square root of 1 plus straight K squared over straight R squared end root equals square root of fraction numerator straight K squared plus straight R squared over denominator straight R squared end fraction end root$