A body of mass M = 100 ; g is rotating with a | vedclass

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(B) The forces acting on the mass are the tension $T$ in the string and the gravitational force $mg$. The horizontal component of tension provides the

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$	an^{-1} \frac{8}{5}$

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(B) The forces acting on the mass are the tension $T$ in the string and the gravitational force $mg$. The horizontal component of tension provides the necessary centripetal force.$T \sin 	heta = m \omega^2 R$$T \cos 	heta = mg$Dividing the two equations,we get:$	an 	heta = \frac{\omega^2 R}{g}$Given frequency $n = 2/\pi \; 	ext{rev/sec}$,so angular velocity $\omega = 2 \pi n = 2 \pi (2/\pi) = 4 \; 	ext{rad/sec}$.Assuming the radius $R = 1 \; m$ (as implied by the calculation in the provided solution snippet):$	an 	heta = \frac{(4)^2 	imes 1}{10} = \frac{16}{10} = \frac{8}{5}$Therefore,$	heta = 	an^{-1} \frac{8}{5}$.

$A$ body of mass $M = 100 \; g$ is rotating with an angular frequency of $2/\pi \; \text{rev/sec}$ as shown in the figure. What is the angle made by the string with the vertical? $(g = 10 \; m/s^2)$

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