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Question

(A) The tension $T$ in a string for a mass $m$ moving in a horizontal circle of radius $r$ with angular velocity $\omega$ is given by the centripetal

Vedclass · Accepted Answer

$\frac{1}{3} \ cycle/s$

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(A) The tension $T$ in a string for a mass $m$ moving in a horizontal circle of radius $r$ with angular velocity $\omega$ is given by the centripetal force: $T = m r \omega^2$.Since $m$ and $r$ are constant,we have $T \propto \omega^2$.Let the initial angular velocity be $\omega_1 = 10 \ cycle/min = \frac{10}{60} \ cycle/s = \frac{1}{6} \ cycle/s$.Let the initial tension be $T_1$ and the final tension be $T_2 = 4T_1$.Using the proportionality $T \propto \omega^2$,we get $\frac{T_2}{T_1} = \left( \frac{\omega_2}{\omega_1} ight)^2$.Substituting the values: $4 = \left( \frac{\omega_2}{\omega_1} ight)^2$,which implies $\frac{\omega_2}{\omega_1} = 2$.Therefore,$\omega_2 = 2 \omega_1 = 2 	imes \frac{1}{6} \ cycle/s = \frac{1}{3} \ cycle/s$.

$A$ mass of $0.5 \ kg$ is attached to a string moving in a horizontal circle with an angular velocity of $10 \ cycle/min$. Keeping the radius constant,the tension in the string is made $4$ times by increasing the angular velocity to $\omega$. The value of $\omega$ for that mass will be:

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