Problem 10 - College Physics

Question

A solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

Step-by-Step Solution

Answer

The normal force must be approximately 5.62 N.

Step 1: Identifying Given Values

First, we need to identify and list all the given values. We have the mass of the cylinder, \( m = 8.25 \, \text{kg} \). The radius of the cylinder is half the diameter, so \( r = \frac{15.0 \, \text{cm}}{2} = 7.5 \, \text{cm} = 0.075 \, \text{m} \). The initial angular velocity is \( \omega_i = 220 \, \text{rpm} \). We need to convert this to radians per second: \[ \omega_i = 220 \, \text{rpm} \times \frac{2\pi}{60} \, \text{rad/s} \approx 23.04 \, \text{rad/s}. \] The number of revolutions the cylinder makes before stopping is 5.25, which can be converted to radians: \[ \theta = 5.25 \, \text{revolutions} \times 2\pi \, \text{rad/revolution} = 10.5\pi \, \text{rad}. \] The coefficient of kinetic friction is \( \mu_k = 0.333 \).

Step 2: Calculating Moment of Inertia

The moment of inertia \( I \) for a solid cylinder is given by \[ I = \frac{1}{2} m r^2. \] Substituting the known values gives us: \[ I = \frac{1}{2} \times 8.25 \, \text{kg} \times (0.075 \, \text{m})^2 \approx 0.0232 \, \text{kg m}^2. \]

Step 3: Determining Angular Deceleration

We use the kinematic equation \( \omega_f^2 = \omega_i^2 + 2\alpha \theta \) where \( \omega_f = 0 \, \text{rad/s} \) (since the cylinder comes to rest) to solve for the angular deceleration \( \alpha \): \[ 0 = (23.04 \, \text{rad/s})^2 + 2\alpha(10.5\pi) \] \[ \alpha = -\frac{(23.04)^2}{2 \times 10.5\pi} \approx -8.06 \, \text{rad/s}^2. \]

Step 4: Relating Torque and Frictional Force

The torque due to friction \( \tau_f \) is related to the frictional force by \( \tau_f = f_r r \), where \( f_r = \mu_k N \). This torque is also related to the moment of inertia and angular acceleration: \( \tau_f = I\alpha \). Equating the two expressions gives: \[ \mu_k N r = I\alpha \] \[ 0.333 N (0.075) = 0.0232 (-8.06). \]

Step 5: Solving for Normal Force

Substitute the known values to solve for \( N \): \[ 0.0333N = 0.0232 \times 8.06 \] \[ N = \frac{0.0232 \times 8.06}{0.0333} \approx 5.62 \, \text{N}. \]

Key Concepts

Understanding Angular Velocity

Angular velocity is a fundamental concept in rotational motion, describing how fast an object rotates. It is represented by the symbol \( \omega \) and usually measured in radians per second \( \text{s}^{-1} \). Think of it as the rotational counterpart to linear velocity.
- For objects moving in a circle, angular velocity tells us the angle that the object sweeps out in a given time.
- The formula to calculate angular velocity is \( \omega = \frac{\theta}{t} \), where \( \theta \) is the angle rotated and \( t \) is time.
For the given exercise, first, we convert angular velocity from revolutions per minute (rpm) to radians per second. Since 1 revolution corresponds to \( 2 \pi \) radians, this conversion helps in seamlessly applying rotational equations.
Understanding angular velocity is crucial, as it influences calculations for other rotational dynamics concepts like moment of inertia and angular deceleration.

Exploring the Moment of Inertia

The moment of inertia, often symbolized by \( I \), is a measure of an object's resistance to changes in its rotational motion. It's comparable to mass in linear motion and reflects how the mass distribution affects rotational stability.
- For a solid cylinder, the moment of inertia is calculated as \( I = \frac{1}{2} m r^2 \), where \( m \) is mass and \( r \) is the radius.
- This exercise highlights the role of geometry in determining rotational dynamics, as the mass distribution (radius and shape) directly influences \( I \).
In our scenario, the given cylinder's moment of inertia helps determine how much torque is needed to change its rotational motion, especially when calculating the frictional force necessary to halt its spin.

Insights into Friction Torque

Friction torque is the resisting torque experienced by a rotating object due to frictional forces. It's a vital concept in bringing rotating systems to a halt or altering their motion.
- Mathematically, it is expressed as \( \tau_f = f_r r \), where \( \tau_f \) is torque, \( f_r \) is the frictional force (given by \( f_r = \mu_k N \)), and \( r \) is the radius.
- The friction torque acts opposite to the direction of rotation, effectively decelerating and eventually stopping the object.
In the exercise, frictional forces are applied by pressing a brake against the cylinder, creating the torque required to reduce and negate angular velocity. Calculating the necessary normal force ensures that the torque applied balances the motion precisely to halt the cylinder.

Deciphering Angular Deceleration

Angular deceleration is the rate of change of angular velocity, indicating how quickly an object is slowing down its rotation. It is represented by \( \alpha \) and measured in radians per second squared \( \text{s}^{-2} \).
- It's analogous to linear deceleration but in the context of circular or rotational motion.
- Angular deceleration can be calculated using kinematic equations, such as \( \omega_f^2 = \omega_i^2 + 2\alpha \theta \).
In our task, knowing the cylinder's initial and final angular velocities (ending at zero, since it stops) and the angular displacement allows us to solve for \( \alpha \). Understanding this concept makes it easier to evaluate how forces interact to alter the spin of a rotating object effectively and predictably.