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(C) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards.Let the centre of the horizontal

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(C) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards.Let the centre of the horizontal circle be $O$. The position vector $\vec{r}$ of the ball relative to $O$ lies in the horizontal plane.The gravitational force $\vec{F}_g = m\vec{g}$ acts vertically downwards through the ball.The torque $\vec{	au}$ about point $O$ is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Since the force of gravity $\vec{F}_g$ is parallel to the vertical axis passing through $O$, and the position vector $\vec{r}$ is horizontal, the torque due to gravity is $\vec{	au}_g = \vec{r} 	imes (m\vec{g})$. The magnitude is $mgr \sin(90^\circ) = mgr$, but the direction is tangential to the circle.However, the tension force $\vec{T}$ acts along the string towards the point of suspension. The line of action of the tension force passes through the point of suspension, but not through $O$.Crucially, for a particle moving in a horizontal circle with constant angular velocity, the net torque about the centre $O$ must be zero because the angular momentum $\vec{L} = \vec{r} 	imes \vec{p}$ is constant in magnitude and direction (the vector $\vec{L}$ points along the vertical axis). Since $\frac{d\vec{L}}{dt} = \vec{	au}_{net} = 0$, the net torque about $O$ is zero.

$A$ small ball of mass $m$ is suspended from the ceiling by a string of length $L$. The ball moves along a horizontal circle with constant angular velocity $\omega$, as shown in the figure. The torque about the centre $(O)$ of the horizontal circle is

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