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(B) For cylinder $P$ (rolling without slipping):By the principle of conservation of energy,the initial potential energy is converted into translationa

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$\sqrt{\frac{3}{2}}$

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(B) For cylinder $P$ (rolling without slipping):By the principle of conservation of energy,the initial potential energy is converted into translational and rotational kinetic energy at the bottom:$mgh = \frac{1}{2}mv_p^2 + \frac{1}{2}I\omega^2$Since $I = \frac{1}{2}mr^2$ and $v_p = r\omega$,we have:$mgh = \frac{1}{2}mv_p^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v_p}{r})^2$$mgh = \frac{1}{2}mv_p^2 + \frac{1}{4}mv_p^2 = \frac{3}{4}mv_p^2$$v_p = \sqrt{\frac{4gh}{3}}$For cylinder $Q$ (sliding without friction):Since there is no friction,the cylinder does not rotate. All potential energy is converted into translational kinetic energy:$mgh = \frac{1}{2}mv_q^2$$v_q = \sqrt{2gh}$The ratio of the speeds is:$\frac{v_q}{v_p} = \frac{\sqrt{2gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{2 \cdot \frac{3}{4}} = \sqrt{\frac{3}{2}}$

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