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(A) According to the law of conservation of energy,the potential energy at the top is converted into translational and rotational kinetic energy at th

Vedclass · Accepted Answer

$v_1 = v_2$

Vedclass · Answer

(A) According to the law of conservation of energy,the potential energy at the top is converted into translational and rotational kinetic energy at the bottom.For a disc of mass $M$ and radius $R$ rolling down a height $h$:$Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$Since the disc rolls without slipping,$\omega = v/R$ and $I = \frac{1}{2} M R^2$.$Mgh = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{1}{2} M R^2) (v/R)^2 = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$$v = \sqrt{\frac{4gh}{3}}$Since both discs start from the same height $h$ and are identical,their velocities at the bottom of the respective inclined planes will be equal,regardless of the length of the incline.Therefore,$v_1 = v_2$.

Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively,as shown in the figure. The velocities of the two discs at the points $A$ and $B$ at the bottom of the inclined planes are related as:

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