A uniform solid cylinder of mass M and radius | vedclass

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(A) The initial rotational kinetic energy of the cylinder is $K_i = \frac{1}{2} I \omega_0^2 = \frac{1}{2} (\frac{1}{2} M R^2) \omega_0^2 = \frac{1}{4

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$\sqrt {\frac{{8{\pi ^2}K}}{M}} $

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(A) The initial rotational kinetic energy of the cylinder is $K_i = \frac{1}{2} I \omega_0^2 = \frac{1}{2} (\frac{1}{2} M R^2) \omega_0^2 = \frac{1}{4} M R^2 \omega_0^2$.When the cylinder rotates by an angle $	heta = 2\pi$,the length of the string wound around the cylinder is $x = R 	heta = R(2\pi) = 2\pi R$.The elastic potential energy stored in the string when it is stretched by length $x$ is $U = \frac{1}{2} K x^2 = \frac{1}{2} K (2\pi R)^2 = \frac{1}{2} K (4\pi^2 R^2) = 2\pi^2 K R^2$.By the law of conservation of energy,the initial kinetic energy must equal the potential energy stored in the string at the point where the cylinder momentarily stops (at $	heta = 2\pi$):$\frac{1}{4} M R^2 \omega_0^2 = 2\pi^2 K R^2$.Solving for $\omega_0$:$\omega_0^2 = \frac{2 \pi^2 K R^2 	imes 4}{M R^2} = \frac{8 \pi^2 K}{M}$.Therefore,$\omega_0 = \sqrt{\frac{8 \pi^2 K}{M}}$.

Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

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The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$

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