Problem 61 - University Physics with Modern 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 required normal force is approximately 7.51 N.

Step 1: Convert Units

First, convert the diameter of the cylinder to radius in meters. The diameter is given as 15.0 cm, so the radius is \( r = \frac{15.0}{2} \times 10^{-2} = 0.075 \) meters. Also, convert the angular velocity from revolutions per minute (rpm) to radians per second (rad/s). The initial angular velocity \( \omega_0 \) is \( 220 \times \frac{2\pi}{60} \approx 23.04 \) rad/s.

Step 2: Determine Total Revolutions in Radians

Convert the total number of turns (5.25 revolutions) into radians. Since 1 revolution equals \(2\pi\) radians, the total rotation in radians is \( 5.25 \times 2\pi \approx 32.99 \) radians.

Step 3: Use Kinematic Equation for Angular Motion

Use the kinematic equation for angular motion: \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Since the final angular velocity \( \omega = 0 \), solve for \( \alpha \) first using \( \omega^2 = \omega_0^2 + 2\alpha\theta \). Plug in \( 0 = (23.04)^2 + 2\alpha(32.99) \) which simplifies to \( \alpha \approx -8.06 \) rad/s².

Step 4: Calculate Torque Due to Friction

The torque \( \tau \) due to friction is defined by \( \tau = I \alpha \), where \( I = \frac{1}{2} m r^2 \) is the moment of inertia for a cylinder. Substitute \( m = 8.25 \text{ kg} \) and \( r = 0.075 \text{ m} \) into \( I \). We get \( I = \frac{1}{2} \times 8.25 \times (0.075)^2 \approx 0.0232 \) kg m². Then, substitute \( \alpha \approx -8.06 \) to find \( \tau = 0.0232 \times (-8.06) \approx -0.187 \) N m.

Step 5: Calculate Required Normal Force

The frictional torque \( \tau \) is also given by \( \tau = rF_f = r\mu F_n \), where \( F_f \) is the frictional force and \( \mu F_n \) is the frictional force expressed in terms of the normal force \( F_n \) and the coefficient of kinetic friction \( \mu = 0.333 \). Substitute \( \tau = -0.187 \text{ N m} \) and solve for \( F_n \), \( F_n = \frac{-\tau}{r\mu} = \frac{-(-0.187)}{0.075 \times 0.333} \approx 7.51 \) N.

Key Concepts

Angular Velocity

Angular velocity measures how fast an object spins or rotates around an axis. It's denoted by \( \omega \) and is typically measured in radians per second (rad/s). To convert from revolutions per minute (rpm) to radians per second, such as the 220 rpm in this exercise, you can use the conversion factor \( \frac{2\pi}{60} \). This is because one revolution is \( 2\pi \) radians, and there are 60 seconds in a minute. Angular velocity is crucial in understanding and predicting rotational motion dynamics because it can show how quickly an object reaches a particular rotational position.

• Key Concept: Initial angular velocity \( \omega_0 = 23.04 \) rad/s
• Relevance: Used to predict when and how a spinning object will stop.

Kinetic Friction

Kinetic friction is the resistive force that occurs between two objects sliding against each other. It is characterized by the coefficient of kinetic friction, \( \mu \), which depends on the materials in contact. In this exercise, the coefficient, \( \mu = 0.333 \), plays a critical role in determining the friction needed to stop the rotating cylinder.Kinetic friction opposes the motion and involves forces parallel to the surfaces in contact. When applying kinetic friction to bring a rotating object to rest, it results in a torque that acts to decelerate the object.

• Role in Problem: Provides the necessary torque to decelerate the rotating cylinder.
• Interaction: Works with the normal force \( F_n \) to create the frictional force \( F_f \).

Moment of Inertia

The moment of inertia, \( I \), is a measure of an object's resistance to changes in its rotation. For a solid cylinder, it depends on its mass and radius, and is computed using the formula \( I = \frac{1}{2} mr^2 \). The moment of inertia provides crucial information about how mass is distributed relative to the axis of rotation. In this exercise, the moment of inertia is used to calculate the torque required to stop the spinning cylinder. With the given mass of 8.25 kg and radius of 0.075 m, the moment of inertia for the cylinder is calculated as \( 0.0232 \mathring{\text{kg}} \text{m}^2 \).

• Importance: Integral in determining how much torque is needed to achieve a certain angular acceleration \( \alpha \).
• Direct Influence: Directly affects the resistance to changes in the cylinder’s rotational speed.