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(B) Let the initial kinetic energy be $E_1 = E$ and the final kinetic energy be $E_2 = 3E$. Since $E = \frac{1}{2}mv^2 = \frac{1}{2}mr^2\omega^2$,we h

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$\frac{E}{3 \pi rm}$

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(B) Let the initial kinetic energy be $E_1 = E$ and the final kinetic energy be $E_2 = 3E$. Since $E = \frac{1}{2}mv^2 = \frac{1}{2}mr^2\omega^2$,we have $E \propto \omega^2$. Thus,$\frac{E_2}{E_1} = \frac{\omega_2^2}{\omega_1^2} = 3$,which implies $\omega_2^2 = 3\omega_1^2$. Assuming the particle starts from rest,$\omega_1 = 0$. However,the problem implies motion starts from an initial state. Let initial angular velocity be $\omega_0$. Then $\omega_f^2 = 3\omega_0^2$. Using the rotational kinematic equation $\omega_f^2 = \omega_0^2 + 2\alpha	heta$,where $	heta = 3 	imes 2\pi = 6\pi$ radians. $3\omega_0^2 = \omega_0^2 + 2\alpha(6\pi) \implies 2\omega_0^2 = 12\alpha\pi \implies \alpha = \frac{\omega_0^2}{6\pi}$. Given $E = \frac{1}{2}mr^2\omega_0^2$,we have $\omega_0^2 = \frac{2E}{mr^2}$. Substituting $\omega_0^2$ into the expression for $\alpha$: $\alpha = \frac{2E}{mr^2} \cdot \frac{1}{6\pi} = \frac{E}{3\pi mr^2}$. The tangential acceleration is $a_t = r\alpha = r \cdot \frac{E}{3\pi mr^2} = \frac{E}{3\pi mr}$.

$A$ particle of mass $m$ moves along a circle of radius $r$ with constant tangential acceleration. If the kinetic energy $E$ of the particle becomes three times by the end of the third revolution after the beginning of the motion,then the magnitude of the tangential acceleration is:

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