Question

The flywheel of an old steam engine is a solid homogeneous metal disk of mass \(M=120 . \mathrm{kg}\) and radius \(R=80.0 \mathrm{~cm} .\) The engine rotates the wheel at \(500 .\) rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force \(F=100 .\) N. If the coefficient of kinetic friction between the pad and the flywheel is \(\mu_{\mathrm{k}}=0.200\), how many revolutions does the flywheel make before coming to rest? How long does it take for the flywheel to come to rest? Calculate the work done by the torque during this time.

Short answer

Answer: The flywheel makes 522 revolutions before coming to rest, it takes 125.57 seconds to stop, and the work done by the torque during this time is 65,536 J.

Step-by-step solution

Initial angular velocity

First, we need to find the initial angular velocity. Given, the engine rotates the wheel at 500 rpm (revolutions per minute). We can convert this to radians per second: \(\omega_0=500 \times \frac{2\pi \,\text{rad}}{60\, \text{s}}= \frac{500}{60} (2\pi) = 52.36\, \text{rad/s}\)

Torque calculation

Next, we will calculate the torque due to the force acting on the flywheel. We are given that the force is radially inward, and the coefficient of kinetic friction is \(\mu_k = 0.20\). The torque is given by: \(\tau = \mu_k F R = (0.20)(100\text{ N})(0.80\text{ m}) = 16\text{ Nm}\)

Moment of inertia and angular acceleration

Now we'll find the moment of inertia of a solid homogeneous disk, which is given by: \(I = \frac{1}{2}MR^2\) Using the given mass and radius, \(I = \frac{1}{2} (120\text{ kg}) (0.80\text{ m})^2 = 38.4\, \text{kg}\cdot\text{m}^2\) Now, let's find the angular acceleration using the torque and moment of inertia: \(\alpha = \frac{\tau}{I} = \frac{16\text{ Nm}}{38.4\text{ kg}\cdot\text{m}^2} = 0.4167\,\text{rad/s}^2\)

Time to stop

Using angular acceleration, we can calculate the time it takes for the flywheel to come to a stop: \(\omega = \omega_0 - \alpha t\) Since the flywheel comes to rest, \(\omega = 0\). Solving for time, \(t = \frac{\omega_0}{\alpha} = \frac{52.36\,\text{rad/s}}{0.4167\,\text{rad/s}^2} = 125.57\,\text{s}\)

Number of revolutions

To find the number of revolutions before coming to rest, we'll use one of the angular kinematic equations: \(\theta = \omega_0 t - \frac{1}{2} \alpha t^2\) Plugging our values and solving: \(\theta = (52.36\,\text{rad/s})(125.57\,\text{s}) - \frac{1}{2} (0.4167\,\text{rad/s}^2)(125.57\,\text{s})^2 = 3280\,\text{rad}\) Converting radians to revolutions: \(\text{revolutions} = \frac{\theta}{2\pi} = \frac{3280\,\text{rad}}{2\pi} = 522\,\text{rev}\)

Work done by torque

Finally, we'll find the work done by the torque during this time. The work done by the torque is the change in kinetic energy of the flywheel: \(W = ΔK = \frac{1}{2}I\omega_0^2 - \frac{1}{2}I\omega^2\) Here, \(\omega = 0\). So, the work done is \(W = \frac{1}{2}(38.4\,\text{kg}\cdot\text{m}^2) (52.36\,\text{rad/s})^2 = 65,536\,\text{J}\) So, the flywheel makes 522 revolutions before coming to rest, it takes 125.57 seconds to stop, and the work done by the torque during this time is 65,536 J.

Key concepts

Angular Velocity

Angular velocity, represented by the symbol \(\omega\), is a measure of the rate at which an object rotates around a particular axis. In simpler terms, it tells us how fast the object is spinning. Just like how we describe the speed of a car in miles per hour, the angular velocity is often described in rotations per minute (rpm) or radians per second (rad/s).

An object spinning faster has a higher angular velocity. It's important to convert angular velocity to a consistent unit, like radians per second, when calculating, as seen in the textbook example where the engine's rpm was converted to rad/s before other calculations were performed. Knowing the angular velocity helps us understand the dynamic behavior of rotating systems and is crucial for solving rotational motion problems.

Torque Calculation

Torque is a measure of the force causing an object to rotate, and it's determined by both the magnitude of the force and the distance from the axis of rotation at which it's applied—the longer the distance (or 'lever arm'), the larger the torque for a given force.

Mathematically, torque is expressed as \(\tau = rF\sin(\theta)\) where \(r\) is the lever arm or radius, \(F\) is the force, and \(\theta\) is the angle between the lever arm and the force vector. In many cases, such as with the flywheel in our problem, the force is applied perpendicularly (a 90-degree angle), simplifying the calculation of torque to just the product of the force and the lever arm radius. In everyday life, when you use a wrench to tighten a bolt, you're applying torque to it.

Moment of Inertia

The moment of inertia, denoted by \(I\), is the rotational equivalent of mass in linear motion. It quantifies an object's resistance to changes in its rotational velocity and depends not only on the mass of the object but also on how that mass is distributed in relation to the axis of rotation.

For a solid homogeneous disk like the flywheel in our textbook problem, the moment of inertia is calculated using the formula \(I = \frac{1}{2}MR^2\). The larger the moment of inertia, the harder it is to start or stop the rotational motion of an object. That's why changing the mass distribution can make a big difference—moving mass closer to or farther from the rotational axis changes the moment of inertia, even if the object's total mass remains the same.

Angular Acceleration

Angular acceleration, symbolized by \(\alpha\), measures the rate of change of angular velocity over time. Just like acceleration in linear motion, it represents how quickly an object speeds up or slows down its rotation. Angular acceleration is a vector quantity, which means it has both magnitude and direction.

The formula to calculate angular acceleration is \(\alpha = \frac{\tau}{I}\), where \(\tau\) is the torque applied to the object and \(I\) is its moment of inertia. In scenarios like the one in the textbook where a brake is applied to a spinning disk, we're concerned with negative angular acceleration, also known as deceleration, which indicates that the object is slowing down.

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by an object due to its rotation and is a component of an object's total kinetic energy. The formula to calculate rotational kinetic energy is \(K = \frac{1}{2}I\omega^2\), where \(I\) is the object's moment of inertia and \(\omega\) is its angular velocity.

This energy plays a central role when we consider the work done by torque as it tells us how much energy is needed to rotate an object up to a certain angular velocity, or conversely, how much energy can be harvested when it slows down, as seen in the solution where the work done was equivalent to the change in rotational kinetic energy of the flywheel. It's analogous to the concept of kinetic energy in linear motion but applied to rotation.