A uniform solid cylinder of mass M = 3 kg an | vedclass

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(A) According to the principle of conservation of energy,the potential energy stored in the spring is converted into the total kinetic energy of the r

Vedclass · Accepted Answer

$1.33$

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(A) According to the principle of conservation of energy,the potential energy stored in the spring is converted into the total kinetic energy of the rolling cylinder (translational + rotational).$\frac{1}{2} kx^2 = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$Since the cylinder rolls without slipping,$\omega = \frac{v}{R}$. For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$.Substituting these into the energy equation:$\frac{1}{2} kx^2 = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$$\frac{1}{2} kx^2 = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$Solving for $v$:$v = \sqrt{\frac{2kx^2}{3M}} = x \sqrt{\frac{2k}{3M}}$Given $M = 3 \ kg$,$k = 8 \ N/m$,and $x = 1 \ m$:$v = 1 	imes \sqrt{\frac{2 	imes 8}{3 	imes 3}} = \sqrt{\frac{16}{9}} = \frac{4}{3} \approx 1.33 \ m/s$.

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