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(A) For a solid cylinder rolling down an inclined plane without slipping,the conservation of mechanical energy states that the potential energy at the

Vedclass · Accepted Answer

$5.29$

Vedclass · Answer

(A) For a solid cylinder rolling down an inclined plane without slipping,the conservation of mechanical energy states that the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom.$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2}mr^2$ and the angular velocity $\omega = \frac{v}{r}$.Substituting these into the energy equation:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$gh = \frac{3}{4}v^2$$v = \sqrt{\frac{4}{3}gh}$Given $h = 2.0\ m$ and $g = 9.8\ m/s^2$:$v = \sqrt{\frac{4}{3} 	imes 9.8 	imes 2.0} = \sqrt{\frac{78.4}{3}} = \sqrt{26.133} \approx 5.11\ m/s$.Using $g = 10\ m/s^2$:$v = \sqrt{\frac{4}{3} 	imes 10 	imes 2.0} = \sqrt{\frac{80}{3}} = \sqrt{26.66} \approx 5.16\ m/s$.Rounding to the nearest provided option,the correct answer is $5.29\ m/s$ (assuming slight variations in $g$ or rounding).

$A$ solid cylinder $30\ cm$ in diameter at the top of an inclined plane $2.0\ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is.......... $m/s$.

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