Question

A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is​

Answer 1

Hello mate : -

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Solution : - In the attached pic

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Thanks

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\large \tt Given \begin{cases} \sf{Mass \: (m) \: = \: 0.5 \: kg} \\ \sf{Radius \: (r) \: = \: 0.5 \: m} \\ \sf{Tension \: or \: Force \: (F) \: = \: 324 \: N}\end{cases}$

Solution :

As we know that :

$\large \star {\boxed{\sf{F \: = \: \dfrac{mv^2}{r}}}} \\ \\ \footnotesize {\pink{\sf{\: \: \: \: \: \: \: \: \: \: \: \: Also \: we \: know \: that \: :}}} \\ \\ \large{\boxed {\sf{v \: = \: r \omega}}} \\ \\ \footnotesize {\underline{\green{\sf{\: \: \: \: \: \: \: \: \: \: Substitute \: value \: of \: v\: \: \: \: \: \: \: \: \: \:}}}} \\ \\ \implies {\sf{F \: = \: \dfrac{m \big( r \omega \big) ^2}{r}}} \\ \\ \implies {\boxed{\sf{F \: = \: m r \omega ^2}}}$

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Put Values in above equation :

$\implies {\sf{324 \: = \: 0.5 \: \times \: 0.5 \: \times \: \omega ^2}} \\ \\ \implies {\sf{\omega ^2 \: = \: \dfrac{324}{0.5 \: \times \: 0.5}}} \\ \\ \implies {\sf{\omega ^2 \: = \: \dfrac{324}{0.25}}} \\ \\ \implies {\sf{\omega ^2 \: = \: 1296}} \\ \\ \implies {\sf{\omega \: = \: \sqrt{1296}}} \\ \\ \implies {\sf{\omega \: = \: 36}} \\ \\ \large {\boxed{\sf{Maximum \: Angular \: Velocity \: is \: 36 \: rad s^{-1}}}}$