Two identical uniform discs roll without slip | vedclass

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Question

(C) The total mechanical energy of a disc rolling without slipping is the sum of its translational and rotational kinetic energies,plus its potential

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$7$

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(C) The total mechanical energy of a disc rolling without slipping is the sum of its translational and rotational kinetic energies,plus its potential energy.Total energy $E = K_{trans} + K_{rot} + U = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + mgh$.For a uniform disc,$I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.Thus,$E = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 + mgh = \frac{3}{4}mv^2 + mgh$.Applying the law of conservation of energy between points $A$ and $B$,and $C$ and $D$:For path $AB$: $E_A = E_B \implies \frac{3}{4}mv_1^2 + mgh_1 = \frac{3}{4}mv_f^2 + 0$,where $h_1 = 30 \ m$.For path $CD$: $E_C = E_D \implies \frac{3}{4}mv_2^2 + mgh_2 = \frac{3}{4}mv_f^2 + 0$,where $h_2 = 27 \ m$.Equating the two expressions for $\frac{3}{4}mv_f^2$: $\frac{3}{4}mv_1^2 + mgh_1 = \frac{3}{4}mv_2^2 + mgh_2$.Substituting the given values: $\frac{3}{4}(3)^2 + 10(30) = \frac{3}{4}v_2^2 + 10(27)$.$\frac{27}{4} + 300 = \frac{3}{4}v_2^2 + 270$.$6.75 + 30 = \frac{3}{4}v_2^2 \implies 36.75 = \frac{3}{4}v_2^2$.$v_2^2 = \frac{36.75 	imes 4}{3} = 12.25 	imes 4 = 49$.$v_2 = \sqrt{49} = 7 \ m/s$.

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