An inextensible string of length ell fixed at | vedclass

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(C) The mass $m$ moves in a horizontal circle of radius $r = \ell \sin 	heta$. The angular frequency is $\omega = 2\pi f = 2\pi 	imes \frac{1}{\pi}

Vedclass · Accepted Answer

$4 m\ell$

Vedclass · Answer

(C) The mass $m$ moves in a horizontal circle of radius $r = \ell \sin 	heta$. The angular frequency is $\omega = 2\pi f = 2\pi 	imes \frac{1}{\pi} = 2 	ext{ rad/s}$. The forces acting on the mass are tension $T$ (along the string) and weight $mg$ (downwards). Resolving tension into components: $T \cos 	heta = mg$ and $T \sin 	heta = m\omega^2 r$. Substituting $r = \ell \sin 	heta$,we get $T \sin 	heta = m\omega^2 \ell \sin 	heta$,which simplifies to $T = m\omega^2 \ell$. Substituting $\omega = 2 	ext{ rad/s}$,we get $T = m(2)^2 \ell = 4m\ell$.

An inextensible string of length $\ell$ fixed at one end,carries a mass $m$ at the other end. If the string makes $\frac{1}{\pi}$ revolutions per second around the vertical axis through the fixed end,the tension in the string is [The string makes an angle $\theta$ with the vertical]

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