Question

A wheel in the form of a uniform disk of radius \(23.0 \mathrm{~cm}\) and mass \(1.40 \mathrm{~kg}\) is turning at 840 rev/min in frictionless bearings. To stop the wheel, a brake pad is pressed against the rim of the wheel with a radially directed force of \(130 \mathrm{~N}\). The wheel makes \(2.80\) revolutions in coming to a stop. Find the coefficient of friction between the brake pad and the rim of the wheel.

Short answer

The coefficient of friction between the brake pad and the rim of the wheel is 0.55.

Step-by-step solution

Convert rotational speed into radians per second

The given rotational speed of the wheel is 840 rev/min. To convert this into radians per second, multiply by \( \frac{2\pi}{60} \), due to 2\(\pi\) radians per revolution and 60 seconds per minute. This gives \( \omega = 840 \times \frac{2\pi}{60} = 88 \, rad/s \) This is the initial angular speed of the wheel.

Find initial angular momentum

The angular momentum of a rotating disk is given by \( L = I\omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed. For a uniform disk, \( I = \frac{1}{2}mR^2 \), where \( m \) is the mass and \( R \) is the radius. Inserting the given values, we find that \( I = \frac{1}{2} \times 1.40 \, kg \times (0.23 m)^2 = 0.0372 \, kg \cdot m^2 \). Therefore, the initial angular momentum is \( L_i = I\omega = 0.0372 \, kg \cdot m^2 \times 88 \, rad/s = 3.27 \, kg \cdot m^2/s \)

Find final angular momentum

Because the wheel comes to a stop, the final angular momentum is zero, \( L_f = 0 \)

Calculate change in angular momentum

The change in angular momentum is simply the initial minus the final, \( \Delta L = L_i - L_f = 3.27 \, kg \cdot m^2/s - 0 = 3.27 \, kg \cdot m^2/s \)

Calculate frictional torque applied by the brake

The formula for angular momentum ties together torque and angular momentum through time: \( \Delta L = \tau \Delta t \), where \( \Delta t \) is the time it takes for the wheel to stop. The problem states the wheel makes 2.80 revolutions in coming to a stop, so the time it takes to stop is \( \Delta t = \frac{2.80 \, rev \times 2\pi \, rad/rev}{\omega} = 0.200 s \). Solving above formula for \( \tau \) and plugging in the known values, we get \( \tau = \frac{\Delta L}{\Delta t} = \frac{3.27 \, kg \cdot m^2/s}{0.200 s} = 16.4 \, N \cdot m \)

Using force to calculate coefficient of friction

The frictional force F causes a torque of \( F \times R \), where \( F \) is the force and \( R \) is the radius, so \( F = \frac{\tau }{R} = \frac{16.4 \, N \cdot m}{0.23 m} = 71.3 \, N \). Since the frictional force is the normal force times the friction coefficient, and the normal force equals the radially directed force which is given as 130 N, you can calculate the friction coefficient \( \mu = \frac{F}{N} = \frac{71.3 \, N}{130 \, N} = 0.55 \)

Key concepts

Angular Momentum

Angular momentum is a fundamental concept in physics, describing the rotational motion of an object. It is closely associated with the rotational equivalent of linear momentum. Mathematically, angular momentum (\( L \)) for a rotating object is the product of its moment of inertia (\( I \)) and its angular velocity (\( \omega \)): \(\), where angular velocity is the rate of change of angular displacement and is measured in radians per second (rad/s).

When a system is closed and not subject to external torques, angular momentum is conserved. In our exercise, we compute initial angular momentum for a spinning wheel to determine how the application of a braking force ultimately affects this conserved quantity. By understanding the relation between torque applied and change in angular momentum, we can explore the effectiveness of the brake.

Moment of Inertia

Moment of inertia, denoted by (\( I \)), signifies the measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution within the object and the axis of rotation. The formula for a uniform solid disk, like the wheel in our exercise, is \(\).

Calculation in Practice

Given the wheel's mass and radius, the moment of inertia is imperative for further calculating the angular momentum and understanding how distributed mass affects angular acceleration. By knowing the moment of inertia, we can also determine how much torque is needed to achieve the desired rotational acceleration or deceleration, illustrating the wheel's stopping action as a direct result of friction.

Coefficient of Friction

The coefficient of friction (\(\)) is the ratio that describes the force of friction between two surfaces in contact. It is a dimensionless value that doesn't have any units. Friction is explained through two main types - static friction, which prevents an object from starting to move, and kinetic friction which opposes motion of a sliding object.

Relevance to the Exercise

By pressing the brake against the wheel, the force of friction arising at the interface helps to halt the wheel's motion. The higher the coefficient of friction, the less force is needed to stop the wheel from turning. Thus, in our exercise, we calculate the coefficient of friction to determine the efficiency of the braking process.

Torque

Torque, often symbolized by (\( \tau \) ), is the measure of the force that can cause an object to rotate about an axis. It is the rotational equivalent of linear force. The equation for torque is \(\), where (\( F \) ) is the magnitude of the force applied, (\( R \) ) is the lever arm or the distance from the axis of rotation to the point where the force is applied and \(\theta\) is the angle between the force vector and the lever arm.

Connection with the Exercise

In our scenario, the brake pad applies a force at the rim of the wheel generating a torque, which in turn decreases the wheel's angular momentum to zero. Calculating this torque allows us to understand the dynamics of the system and how forces can be efficiently applied to control rotational motion.