A ball of mass m is attached to the free end | vedclass

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(B) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards. The tension $T$ can be resolved

Vedclass · Accepted Answer

$\sqrt{\frac{T}{m l}}$

Vedclass · Answer

(B) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards. The tension $T$ can be resolved into two components: $T \cos 	heta$ acting vertically upwards and $T \sin 	heta$ acting horizontally towards the center of the circular path.For vertical equilibrium:$T \cos 	heta = mg$  --- $(1)$For horizontal circular motion,the centripetal force is provided by the horizontal component of tension:$T \sin 	heta = mr \omega^2$  --- $(2)$From the geometry of the figure,the radius of the circular path is $r = l \sin 	heta$.Substituting $r$ into equation $(2)$:$T \sin 	heta = m(l \sin 	heta) \omega^2$Dividing both sides by $m l \sin 	heta$ (assuming $\sin 	heta 
eq 0$):$\omega^2 = \frac{T}{ml}$Therefore,the angular velocity is:$\omega = \sqrt{\frac{T}{ml}}$

$A$ ball of mass $m$ is attached to the free end of a string of length $l$. The ball is moving in a horizontal circular path about the vertical axis as shown in the diagram. The angular velocity $\omega$ of the ball will be ($T =$ Tension in the string).

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