Question

A \(0.0500-\mathrm{kg}\) bead slides on a wire bent into a circle of radius \(0.400 \mathrm{~m}\) You pluck the bead with a force tangent to the circle. What force is needed to give the bead an angular acceleration of \(6.00 \mathrm{rad} / \mathrm{s}^{2} ?\)

Short answer

Answer: The tangential force needed to give the bead an angular acceleration of 6.00 rad/s² is 1.20 N.

Step-by-step solution

Identify the known variables and required force

We are given the mass of the bead \(m = 0.0500\,\text{kg}\), the radius of the wire circle \(r = 0.400\,\text{m}\), and the desired angular acceleration \(\alpha = 6.00\,\text{rad/s}^2\). We need to find the force \(F\) needed to achieve this angular acceleration.

Write down the relationship between torque, moment of inertia, and angular acceleration

We know that torque \(\tau = I\alpha\), where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. The moment of inertia for a point mass rotating about an axis is given by \(I = mr^2\).

Substitute moment of inertia equation into torque equation

We substitute the equation of moment of inertia, \(I = mr^2\), into the torque equation: \(\tau = (mr^2)\alpha\).

Find the torque needed to achieve the angular acceleration

Now we can find the desired torque using the given values for mass, radius, and angular acceleration: \(\tau = (0.0500\,\text{kg})(0.400\,\text{m})^2(6.00\,\text{rad/s}^2) = 0.48\, \text{N}\cdot\text{m}\).

Find the tangential force required

The tangential force \(F\) acts at a distance \(r\) from the rotation axis, thus creating torque \(\tau\). So, \(\tau = Fr\). Now we will solve for \(F\): \(F = \frac{\tau}{r}\).

Calculate the force needed to achieve the angular acceleration

Using the values for torque and radius, we have \(F = \frac{0.48\, \text{N}\cdot\text{m}}{0.400\,\text{m}} = 1.20\, \text{N}\). Thus, a tangential force of \(1.20\,\text{N}\) is needed to give the bead an angular acceleration of \(6.00\,\text{rad/s}^2\).

Key concepts

Torque

Understanding how forces cause rotation is crucial in mechanics. Torque, often symbolized by \( \tau \), is the manifestation of a force's ability to rotate an object around an axis. Think of torque as the twisting force applied to a doorknob; it's what makes the door rotate on its hinges.

When we apply a tangential force to rotate an object like a bead on a wire, we're essentially generating torque. The formula \( \tau = I\alpha \) gives us the relationship between torque (\( \tau \) being the torque), moment of inertia (\( I \) representing the moment of inertia), and the angular acceleration (\( \alpha \) being the angular acceleration). An important thing to remember is that the torque produced is proportional to the force applied and the distance from the axis of rotation, encapsulated in the simple formula \( \tau = Fr \) where \( F \) is the applied force and \( r \) is the distance from the axis.

Moment of Inertia

The moment of inertia, denoted by \( I \), is akin to mass in linear motion, but it's for rotational motion. Instead of resistance to change in speed, \( I \) measures an object's resistance to changes in its rotational speed. You can think of it as the 'rotational mass.'

Each object has a different moment of inertia, depending on its mass distribution with respect to the axis of rotation. For a single point mass like our bead, the moment of inertia is given by the formula \( I = mr^2 \), where \( m \) is the mass and \( r \) is the radial distance to the rotation axis.

In practical terms, the farther the mass is from the rotation axis, the larger the moment of inertia, making it harder to spin the object. In our exercise, we calculate this to find the torque needed to cause the desired angular acceleration.

Tangential Force

A tangential force is the force applied perpendicular to the radius of a circle, which in essence, 'tangentially' touches the circle at the point of application. This force is the real-world effort, such as a push or pull, that instigates rotational motion.

For an object in circular motion, like the bead on the wire, a tangential force causes a change in its rotational speed. The amount of tangential force required depends on the torque needed and the radius of the circle: \( F = \frac{\tau}{r} \), where \( F \) is the force, \( \tau \) is the torque, and \( r \) is the radius.

Applying enough tangential force will allow the bead to achieve the specified angular acceleration. As shown in the exercise, calculating this force helps us understand the relationship between linear forces applied at a distance and their rotational effects.