Question

A 0.050 -kg bead slides on a wire bent into a circle of radius \(0.40 \mathrm{~m}\). You pluck the bead with a force tangent to the circle. What force is needed to give the bead an angular acceleration of \(6.0 \mathrm{rad} / \mathrm{s}^{2} ?\)

Short answer

Answer: The force needed to achieve the given angular acceleration is 0.12 N.

Step-by-step solution

Write down Newton's second law for rotational motion

Newton's second law for rotational motion is given by: \(\tau = I \alpha\) Where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration.

Find the moment of inertia of the bead

Since the bead is moving on a circular path with a constant radius, we can treat it as a point mass and use the formula for the moment of inertia of a point mass: \(I = mr^2\) where \(m\) is the mass of the bead, and \(r\) is the radius of the circle. Plugging in the given values, we get: \(I = (0.050\,\text{kg})(0.40\,\text{m})^2 = 0.008\,\text{kg}\cdot\text{m}^2\)

Calculate the torque needed to achieve the given angular acceleration

Now that we have the moment of inertia, we can use Newton's second law for rotational motion to find the torque needed to achieve the given angular acceleration: \(\tau = I \alpha\) Plugging in the values for \(I\) and \(\alpha\), we get: \(\tau = (0.008\,\text{kg}\cdot\text{m}^2)(6.0\,\text{rad}/\text{s}^2) = 0.048\,\text{N}\cdot\text{m}\)

Determine the tangential force needed to produce the required torque

Since the force is applied tangentially to the bead, the torque produced by the force can be calculated as: \(\tau = r F_\text{t}\) where \(F_\text{t}\) is the tangential force. We can now solve for \(F_\text{t}\): \(F_\text{t} = \frac{\tau}{r}\) Substituting the values, we get: \(F_\text{t} = \frac{0.048\,\text{N}\cdot\text{m}}{0.40\,\text{m}} = 0.12\,\text{N}\) Therefore, the force needed to give the bead an angular acceleration of \(6.0\,\text{rad}/\text{s}^2\) is \(0.12\,\text{N}\).

Key concepts

Newton's Second Law for Rotational Motion

Newton's Second Law for Rotational Motion is a pivotal concept that links the rotational counterparts of force, mass, and acceleration from linear motion. In rotational dynamics, the law is expressed as \(\tau = I \alpha\), where \(\tau\) represents the torque, \(I\) is the moment of inertia, and \(\alpha\) denotes the angular acceleration. This equation is quite similar to the linear form \(F = ma\), where \(F\) is force, \(m\) is mass, and \(a\) is linear acceleration. In essence, Newton's second law suggests that the angular acceleration of an object is directly proportional to the net torque acting on it and inversely proportional to its moment of inertia.

To understand this further, if you apply a larger torque, the angular acceleration increases. Meanwhile, a higher moment of inertia, which is like rotational resistance, makes it harder to accelerate the object rotationally. Imagine rotating a door; the closer you apply the force to the hinges, the more challenging it is to achieve the same rotational effect, because the door's moment of inertia is higher near the hinge. This concept is crucial in understanding how objects behave when forces are applied away from their center of mass.

Moment of Inertia

The Moment of Inertia is a measure of an object's resistance to changes in its rotational motion. It depends not only on the object's mass but also on how that mass is distributed relative to the axis of rotation. For the bead moving on a wire circle, it's treated as a point mass. The formula for the moment of inertia \(I\) of a point mass is \(I = mr^2\), where \(m\) is the mass and \(r\) is the radius, or distance from the axis.

This distribution aspect is crucial because it means identical masses can have different moments of inertia if they are at different distances from the axis. For instance, if you hold a dumbbell with weights at either end and rotate it around a central axis, it will be harder to start or stop this motion compared to if the weights were near the center. The further the mass, the more torque is needed to change its rotational state, essentially because it has a larger moment of inertia.

• Choosing different axis points for rotation changes the moment of inertia.
• A higher moment of inertia implies more force is required for the same rotational change.
• This concept helps in designing objects like flywheels and disks, where rotational stability is essential.

Understanding the moment of inertia aids in predicting how an object will behave when subjected to rotational forces.

Torque Calculation

Torque calculation bridges the gap between linear forces and angular motion. Torque \(\tau\) is the rotational equivalent of force, essential in changing an object's rotational state. It's calculated as \(\tau = r F_t\), where \(r\) is the distance from the pivot point to the point where the force is applied, and \(F_t\) is the tangential component of the force.

For the given exercise, after determining that the required torque is 0.048 N·m to achieve the specific angular acceleration, we need to find the respective force acting tangentially. By rearranging the torque formula to \(F_t = \frac{\tau}{r}\), and inputting the bead's radius, it's found that the tangential force is 0.12 N. This force is what you'd need to apply to pluck the bead and set it rotating with the required angular acceleration.

Important points about torque and force include:

• Torque is larger when the force is applied further from the axis of rotation.
• Tangential force contributes directly to the rotation, perpendicular to the radius.
• Unit consistency is crucial; ensure all measurements use appropriate units, like meters and newtons, for coherent results.

Understanding torque and how to calculate it helps in dealing with multiple physics problems involving rotational dynamics effectively.