A disc is rolling on an inclined plane without slipping then what fraction of its total energy will be in form of rotational kinetic energy :-

Question

A disc is rolling on an inclined plane without slipping then what fraction of its total energy will be in form of rotational kinetic energy :- (A) 1: 3 (B) 1: 2 (C) 2: 7 (D) 2: 5

Tardigrade Admin · Accepted Answer

Fraction =(K text Rotation /K text Total )=((K2/R2)/1+(K2)R2) For disc, (K2/R2)=(1/2) so, fraction =((1/2)/1+(1)2) =1: 3

Q. A disc is rolling on an inclined plane without slipping then what fraction of its total energy will be in form of rotational kinetic energy :-

$1 : 3$

$1 : 2$

$2 : 7$

$2 : 5$

Fraction $= \frac{K _{Rotation}}{K _{Total}} = \frac{\frac{K ^{2}}{R ^{2}}}{1 + \frac{K ^{2}}{R ^{2}}}$
For disc, $\frac{K ^{2}}{R ^{2}} = \frac{1}{2}$
so, fraction $= \frac{\frac{1}{2}}{1 + \frac{1}{2}}$
$= 1 : 3$