Problem 74 - College Physics

Question

While repairing his bicycle, a student turns it upside down and sets the front wheel spinning at 2.00 rev \(/ \mathrm{s}\). Assume the wheel has a mass of \(3.25 \mathrm{~kg}\) and all of the mass is located on the rim, which has a radius of \(41.0 \mathrm{~cm}\). To slow the wheel, he places his hand on the tire, thereby exerting a tangential force of friction on the wheel. It takes \(3.50 \mathrm{~s}\) to come to rest. Use the change in angular momentum to determine the force he exerts on the wheel. Assume the frictional force of the axle is negligible.

Step-by-Step Solution

Answer

The student exerts a force of 4.79 N on the wheel.

Step 1: Find the Angular Velocity

First, we're asked to find the initial angular velocity in radians per second. Given, the wheel spins at 2.00 revolutions per second initially. Convert this to radians per second using the conversion factor \(2\pi \) radians per revolution: \[ \omega_i = 2.00 \text{ rev/s} \times 2\pi \text{ rad/rev} = 4\pi \text{ rad/s} \]

Step 2: Calculate the Moment of Inertia

The problem states that all the mass is located on the rim of the wheel, which suggests that the wheel can be modeled as a hoop. The moment of inertia \(I\) for a hoop is given by \(I = mr^2\), where \(m\) is the mass of the hoop and \(r\) is the radius:\[ I = 3.25 \text{ kg} \times (0.41 \text{ m})^2 = 0.547925 \text{ kg} \cdot \text{m}^2 \]

Step 3: Determine the Angular Deceleration

The wheel comes to rest in 3.50 seconds, so the final angular velocity \( \omega_f = 0 \). The angular deceleration \(\alpha\) is calculated using the formula \( \omega_f = \omega_i + \alpha t \).Solving for \(\alpha\):\[ 0 = 4\pi + \alpha (3.50) \]\[ \alpha = \frac{-4\pi}{3.50} = -\frac{4\pi}{3.50} \text{ rad/s}^2 \]

Step 4: Calculate the Torque

Torque \(\tau\) is related to the moment of inertia \(I\) and angular deceleration \(\alpha\) by \(\tau = I\alpha\). Using the values calculated previously:\[ \tau = 0.547925 \text{ kg} \cdot \text{m}^2 \times \left(-\frac{4\pi}{3.50}\right) \text{ rad/s}^2 \] \[ \tau = -1.963437 \text{ N} \cdot \text{m} \]

Step 5: Find the Force Exerted

The relationship between torque \(\tau\) and the radius \(r\) and force \(F\) is \(\tau = rF\). Solving for the force \(F\):\[ F = \frac{\tau}{r} = \frac{-1.963437 \text{ N} \cdot \text{m}}{0.41 \text{ m}} \]\[ F = -4.788 \text{ N} \] The negative sign indicates direction, but for the magnitude, the force is \(4.788 \text{ N}\).

Step 6: Conclusion

The student exerts a tangential force of 4.79 N on the wheel to bring it to a stop.

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast something is rotating. In the context of a wheel, it tells us how many radians the wheel spins through per unit time—like seconds. This is essential for understanding how quickly a bicycle wheel turns when in motion.
Let's break this down: Angular velocity is often represented by the symbol \( \omega \). To convert from revolutions per second to radians per second, we use the factor that one complete revolution is \( 2\pi \) radians. Therefore, if a wheel spins at 2.00 revolutions per second, its angular velocity in radians per second, \( \omega_i \), is \( 2.00 \times 2\pi \), which results in \( 4\pi \) rad/s.
It's useful to understand this because angular velocity gives us a solid foundation to examine more complex concepts like angular acceleration and torque.

Moment of Inertia

Moment of inertia can be thought of as the rotational equivalent of mass in linear motion. It's a measure of how much an object resists changes in its rotation. The larger the moment of inertia, the harder it is to speed up or slow down the object’s rotation.
In our example, the wheel is modeled as a hoop. This means all its mass is concentrated at its rim. The formula for the moment of inertia \( I \) of a hoop is \( I = mr^2 \), where \( m \) is the mass and \( r \) is the radius. For a wheel with a mass of 3.25 kg and radius of 0.41 meters, the moment of inertia is calculated as \( I = 3.25 \times (0.41)^2 = 0.547925 \; \text{kg} \cdot \text{m}^2 \).
Understanding moment of inertia is crucial because it directly affects how torque is applied to a rotating object.

Angular Deceleration

Angular deceleration is essentially angular acceleration but in the opposite direction. It describes how quickly an object slows down its rotation. In the exercise, the wheel's final angular velocity is 0 rad/s since it comes to rest after 3.50 seconds.
To find angular deceleration, denoted by \( \alpha \), we use the equation: \( \omega_f = \omega_i + \alpha t \). Here, \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity, \( \alpha \) is the angular deceleration, and \( t \) is the time. Substituting the values gives us: \( 0 = 4\pi + \alpha(3.50) \), leading to \( \alpha = \frac{-4\pi}{3.50} \; \text{rad/s}^2 \).
This tells us how the speed of rotation decreases over time, which is vital for calculating the force required to stop the rotation.

Torque

Torque is the force that causes an object to rotate. It’s the rotational equivalent of linear force and plays a crucial role in spinning activities.
In the exercise, torque \( \tau \) is used to describe the effect of the frictional force applied to the stopping wheel. The formula for torque is \( \tau = I\alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular deceleration. Substituting the values, we find \( \tau = 0.547925 \times \left(\frac{-4\pi}{3.50}\right) \; \text{N} \cdot \text{m} \).
Torque can also be related to the force \( F \) exerted at a distance \( r \) from the pivot point, using \( \tau = rF \). By rearranging, you can find the force: \( F = \frac{\tau}{r} \). For our example, it results in a force of \( 4.788 \text{ N} \), indicating how much force the student must exert to bring the wheel to a standstill.