Question

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

• A. mgL sin θ
• B. mgL cos θ
• C. 0
• D. mgl cos θ

Hint:

The torque in a rotational system can be calculated by multiplying the force by the perpendicular distance from the axis of rotation.

Answer 1

The Correct Option is C

1. Step 1: The ball is moving in a horizontal circle, which means that the forces acting on it in the vertical direction (the gravitational force mg) are balanced by the vertical component of the tension in the string. The ball is subject to two forces:
Gravitational force: mg and Tension in the string: T
2. Step 2: Since the ball is in circular motion, the horizontal component of the tension T sin θ provides the centripetal force required for the circular motion. The vertical component T cos θ balances the weight of the ball.
3. Step 3: The torque about the center O of the horizontal circle is given by the cross product of the force and the radius vector. The torque τ due to the forces acting on the ball is calculated as:
τ = r × F
where r is the radius vector (the length L) and F is the force (in this case, the tension). However, since the tension is acting along the string, there is no torque about the center of the horizontal circle.
4. Step 4: Therefore, the total torque about the center is zero, because the force creating the circular motion (tension in the string) does not create any rotational effect about the center of the horizontal circle.

Answer 2

 The Correct answer is: Option 3

The problem asks for the torque about the center (O) of the horizontal circle.

We are given a hint: Angular momentum is conserved about O.

The angular momentum $\vec{L}$ of the ball about point O is given by:

$\vec{L} = \vec{r} \times \vec{p} = m (\vec{r} \times \vec{v})$

where $\vec{r}$ is the position vector from O to the ball, and $\vec{v}$ is the velocity of the ball.

The ball moves in a horizontal circle of radius $r = L \sin\theta$ with a constant angular velocity $\omega$. The linear speed of the ball is $v = r\omega = L \sin\theta \omega$.

Let's set up a coordinate system where O is the origin, the horizontal plane is the xy-plane, and the vertical direction is the z-axis. At a particular instant, let the position of the ball be $\vec{r} = L \sin\theta \hat{i}$. For counter-clockwise motion, the velocity vector $\vec{v} = L \sin\theta \omega \hat{j}$.

The angular momentum $\vec{L}$ about O is:

$\vec{L} = m (\vec{r} \times \vec{v}) = m ((L \sin\theta \hat{i}) \times (L \sin\theta \omega \hat{j}))$

$\vec{L} = m L^2 \sin^2\theta \omega (\hat{i} \times \hat{j})$

$\vec{L} = m L^2 \sin^2\theta \omega \hat{k}$

The angular momentum vector $\vec{L}$ is constant because $m$, $L$, $\theta$, and $\omega$ are all constants.

The relationship between the net torque $\vec{\tau}_{net}$ acting on a system and its angular momentum $\vec{L}$ is given by:

$\vec{\tau}_{net} = \frac{d\vec{L}}{dt}$

Since the angular momentum $\vec{L}$ is constant, its time derivative is zero:

$\vec{\tau}_{net} = \frac{d}{dt} (m L^2 \sin^2\theta \omega \hat{k}) = 0$

Therefore, the net torque about the center (O) of the horizontal circle is zero.