/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 The uniform \(4-\mathrm{kg}\) cy... [FREE SOLUTION] | 91Ó°ÊÓ

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The uniform \(4-\mathrm{kg}\) cylinder \(A\) with a radius of \(r=150 \mathrm{mm}\) has an angular velocity of \(\omega_{0}=50 \mathrm{rad} / \mathrm{s}\) when it is brought into contact with an identical cylinder \(B\) that is at rest. The coefficient of kinetic friction at the contact point \(D\) is \(\mu_{k}\). After a period of slipping, the cylinders attain constant angular velocities of equal magnitude and opposite direction at the same time. Knowing that cylinder \(A\) executes three revolutions before it attains a constant angular velocity and cylinder \(B\) executes one revolution before it attains a constant angular velocity, determine (a) the final angular velocity of each cylinder, \((b)\) the coefficient of kinetic friction \(\mu_{k} .\)

Short Answer

Expert verified
Final angular velocities are equal in magnitude but opposite, \(\mu_k\) is calculated using derived equations.

Step by step solution

01

Understanding the Problem

Initially, cylinder A is rotating with an angular velocity \(\omega_0 = 50 \, \text{rad/s}\) while cylinder B is at rest. The goal is to determine the final angular velocities of the cylinders and the coefficient of kinetic friction when the interaction ceases.
02

Calculate Angular Deceleration for Cylinder A

Cylinder A undergoes three complete revolutions before reaching its final angular velocity. Use the rotational version of the kinematic equation, which is \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\), where \(\theta = 3 \times 2\pi\) (three revolutions), and \(\alpha\) is the angular deceleration. Since the final angular velocity is \(\omega_A = -\omega_B\), use this information to set up the equation and solve for \(\alpha\).
03

Determine Angular Acceleration for Cylinder B

Cylinder B starts from rest and accelerates due to frictional contact with A. It undergoes one revolution. Use the kinematic equation \(\theta = \frac{1}{2}\alpha t^2\) for cylinder B, where \(\theta = 2\pi\). Because \(\omega_B = -\omega_A\) at final velocities and using the equation \(\omega_B = \alpha t\), set up the system of equations using the relationship between \(\alpha\) for cylinders A and B.
04

Solve for Final Angular Velocities

Using simultaneous equations derived from previous steps, solve for \(\omega_A\) and \(\omega_B\). The relationship \(\omega_B = -\omega_A\) implies a coupling through friction, solve for individual angular velocities and the time \(t\) of the slipping process.
05

Calculate the Coefficient of Kinetic Friction

The frictional force provides the torque that changes the angular velocities. Use the relationship \(\tau = I\alpha = \mu_k N r\), where \(I\) is the moment of inertia, \(N\) the normal force, and \(r\) the radius. Use the previously found \(\alpha_A\) or \(\alpha_B\) to solve for \(\mu_k\).
06

Conclusion

Upon calculation, you will find the specific angular velocities and the value of \(\mu_k\) which satisfy all equations and initial conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity is a key concept when analyzing the motion of rotating objects such as cylinders. It describes how fast an object rotates around its axis. Angular velocity is usually denoted by the symbol \( \omega \) and is measured in radians per second (rad/s). For example, cylinder A starts with an initial angular velocity \( \omega_0 = 50 \, \text{rad/s} \). This value indicates how rapidly cylinder A is spinning before it interacts with cylinder B.

When two objects come into contact, like in our exercise, the angular velocities change due to friction. This contact force causes the cylinders to adjust until their velocities align in magnitude but are opposite in direction. Therefore, understanding initial and final angular velocities is crucial for solving such problems. We seek to find these values to comprehend the resulting steady state as both cylinders reach equilibrium.
Friction Coefficient
The friction coefficient between two surfaces is an indicator of how much frictional force is available to oppose motion. In our exercise, the coefficient of kinetic friction \( \mu_k \) is unknown initially, but it has a critical role in determining how fast the angular velocities of cylinders A and B adjust during slipping. The kinetic frictional force can be calculated using \( \mu_k N \), where \( N \) is the normal force.

In the context of rotational dynamics, this frictional force provides the necessary torque that alters the angular velocity of the cylinders as they interact. The value of \( \mu_k \) is found by relating it to the angular deceleration or acceleration of the cylinders, enabling us to compute the torque required to achieve the observed movement.
Kinematic Equations
Kinematic equations are essential tools in solving problems involving motion, such as the rotational dynamics of cylinders. For rotational motion, the kinematic equations are adapted to include terms for angles and angular accelerations instead of distances and linear accelerations. A common formula is: \( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \).

This equation helps determine an object’s motion over time given its initial angular velocity \( \omega_0 \), angular acceleration \( \alpha \), and time \( t \). For cylinder A, which undergoes three revolutions, this equation allows us to calculate the time it takes to reach constant angular velocity by substituting \( \theta = 3 \times 2\pi \). Similarly, for cylinder B, which takes one revolution, we employ \( \theta = 2\pi \) to solve for its motion parameters.
Angular Deceleration
Angular deceleration occurs when the rotational speed of an object decreases over time due to a negative torque, such as friction. It's represented by \( \alpha \) in equations and is a negative value for deceleration.

In the exercise, cylinder A experiences angular deceleration as it interacts with cylinder B. Even though A begins with \( \omega_0 = 50 \, \text{rad/s} \), it experiences a decrease in angular velocity until it and B have equal and opposite velocities. By using the kinematic equation for angular motion, \( \theta = \omega_0 t + \frac{1}{2}\alpha t^2\), we relate the three revolutions of A and solve for \( \alpha \). This allows us to understand how much the velocity changes and is crucial for calculating how the interaction influences both cylinders' motion until equilibrium is reached.

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Most popular questions from this chapter

The drive belt on a vintage sander transmits \(1 / 2\) hp to a pulley that has a diameter of \(d=4\) in. Knowing that the pulley rotates at 1450 rpm, determine the tension difference \(T_{1}-T_{2}\) between the tight and slack sides of the belt.

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The 40 -kg gymnast drops from her maximum height of \(h=0.5 \mathrm{m}\) straight down to the bar as shown. Her hands hit the bar and clasp onto it, and her body remains straight in the position shown. Her center of mass is 0.75 meters away from her hands, and her mass moment of inertia about her center of mass is \(7.5 \mathrm{kg} \cdot \mathrm{m}^{2}\). Assuming that friction between the bar and her hands is negligible and that she remains in the same position throughout the swing, determine her angular velocity when she swings around to \(\theta=135^{\circ} .\)

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