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

Additionally, what is the work done by the torque during this time? Answer: The flywheel makes approximately 524 revolutions before coming to rest, and it takes approximately 125.9 seconds for the flywheel to stop. The work done by the torque during this time is approximately 52,689.6 J.

Step-by-step solution

Calculate the initial angular velocity

First, we need to convert the given rotation speed in rpm (rotations per minute) to rad/s (radians per second). For 1 rotation, there are 2π radians. So, we have: $$ \omega = 500\,\text{rpm} \times \frac{2\pi\,\text{rad}}{1\,\text{rotation}} \times \frac{1\,\text{min}}{60\,\text{s}} \approx 52.4\,\text{rad/s} $$

Calculate the moment of inertia

Next, we need to find the moment of inertia (I) of the homogeneous metal disk. The moment of inertia of a solid disk with mass M and radius R is given by the formula: $$ I = \frac{1}{2}MR^2 $$ Plugging in the given values, we have: $$ I = \frac{1}{2}(120\,\text{kg})(0.8\,\text{m})^2 = 38.4\,\text{kg}\cdot\text{m}^2 $$

Calculate the torque and angular deceleration

The torque (τ) of the given force (F) and coefficient of kinetic friction (μk) acting at the edge of the flywheel can be calculated using the formula: $$ \tau = \mu_{k}FR $$ Plugging in the given values, we get: $$ \tau = (0.2)(100\,\text{N})(0.8\,\text{m}) = 16\,\text{N}\cdot\text{m} $$ Now we can find the angular deceleration (α) using Newton's second law for rotation: $$ \tau = I\alpha $$ Solving for α: $$ \alpha = \frac{\tau}{I} = \frac{16\,\text{N}\cdot\text{m}}{38.4\,\text{kg}\cdot\text{m}^2} \approx 0.417\,\text{rad/s}^2 $$

Calculate the time taken to stop

Using the kinematic equation for angular motion, we have: $$ \omega_f = \omega_i - \alpha t $$ Since the final angular velocity (ωf) will be zero when the flywheel comes to rest, we can solve for the stopping time (t): $$ t = \frac{\omega_i}{\alpha} = \frac{52.4\,\text{rad/s}}{0.417\,\text{rad/s}^2} \approx 125.9\,\text{s} $$

Calculate the number of revolutions before coming to rest

Now we can calculate the angular displacement (θ) before the flywheel comes to rest using another kinematic equation for angular motion: $$ \theta = \omega_i t - \frac{1}{2}\alpha t^2 $$ Plugging in the values, we get: $$ \theta = (52.4\,\text{rad/s})(125.9\,\text{s}) - \frac{1}{2}(0.417\,\text{rad/s}^2)(125.9\,\text{s})^2 \approx 3293.1\,\text{rad} $$ To find the number of revolutions, we divide the angular displacement by 2π radians per revolution: $$ \text{Revolutions} = \frac{\theta}{2\pi} \approx \frac{3293.1\,\text{rad}}{2\pi\,\text{rad}} \approx 524\,\text{revolutions} $$

Calculate the work done by the torque

The work done by the torque can be calculated using the formula: $$ W = \tau \theta $$ Plugging in the values, we get: $$ W = (16\,\text{N}\cdot\text{m})(3293.1\,\text{rad}) \approx 52,689.6\,\text{J} $$ So, the flywheel makes approximately 524 revolutions before coming to rest, and it takes around 125.9 seconds for the flywheel to come to rest. The work done by the torque during this time is approximately 52,689.6 J.

Key concepts

Moment of Inertia

The moment of inertia, represented by the symbol 'I', is essentially a measure of an object's resistance to changes to its rotation. In the case of the flywheel in our example, its moment of inertia depends on the mass of the flywheel and how this mass is distributed relative to the axis of rotation.

For a solid, homogeneous metal disk like our flywheel, the moment of inertia can be calculated using the formula: \[I = \frac{1}{2}MR^2\] where 'M' is the mass of the disk, and 'R' stands for its radius. So, a larger mass or a larger radius would both result in a greater moment of inertia, meaning it would be harder to alter the flywheel's rotational motion. This is why heavy flywheels can store substantial amounts of rotational energy.

Angular Velocity

Angular velocity is the rate at which an object rotates or spins around an axis and is denoted by the Greek letter omega (\(\omega\)). It is typically measured in radians per second (rad/s).

In our problem, the initial angular velocity of the flywheel was given in revolutions per minute (rpm), which we converted to radians per second. This conversion is important because all subsequent calculations involving rotational motion, such as finding angular displacement or kinetic energy, rely on angular velocity being in the consistent unit of rad/s.

Torque

Torque, represented by the Greek letter tau (\(\tau\)), is the rotational equivalent of a force. It describes the tendency of a force to rotate an object around an axis, pivot, or fulcrum. The magnitude of the torque depends on the force applied, the distance at which the force is applied from the axis of rotation (this distance is called the 'lever arm'), and the angle between the force and the lever arm.

In the context of our flywheel, torque is caused by kinetic friction between the brake pad and the wheel at a certain distance (the radius of the flywheel) from the center. The formula \[\tau = \mu_kFR\]takes into account the coefficient of kinetic friction (\(\mu_k\)), the applied force (F), and the radius (R), to calculate the torque that eventually slows down the flywheel.

Kinetic Friction

Kinetic friction occurs when two surfaces are moving relative to each other and rub together. The coefficient of kinetic friction, denoted as \(\mu_k\), is a measure of how much frictional resistance there is between the surfaces in contact.

It's dimensionless and typically less than the coefficient of static friction (which acts when the surfaces are not moving relative to each other). In our problem, this kinetic friction between the brake pad and the flywheel generates the torque needed to bring the flywheel to a stop. It's important to know that kinetic friction, and therefore torque, is constant in this scenario, provided the conditions of the surfaces do not change.

Work Done By Torque

The concept of work in physics relates to force acting on an object to move it some distance. In rotational motion, work is done by torque (instead of linear force) when it causes an object to rotate around an axis. The calculation of work done by torque (W) is a bit different from linear work but they're analogous. The formula for work done by torque in rotational motion is: \[W = \tau \theta\]where \(\tau\) is the torque and \(\theta\) is the angular displacement in radians. In the process of the flywheel slowing down and eventually coming to a stop, the brake pad performs work against the kinetic energy of the flywheel, eventually draining it to bring the wheel to a rest.