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Chapter 17: Problem 86

The circular platform A is fitted with a rim of 200-mm inner radius and can rotate freely about the vertical shaft. It is known that the platform-rim unit has a mass of 5 kg and a radius of gyration of 175 mm with respect to the shaft. At a time when the platform is rotating with an angular velocity of 50 rpm, a 3-kg disk B of radius 80 mm is placed on the platform with no velocity. Knowing that disk B then slides until it comes to rest relative to the platform against the rim, determine the final angular velocity of the platform.

Short Answer

Expert verified
The final angular velocity of the platform is approximately 28.05 rpm.

Step by step solution

01

Determine Initial Angular Momentum

The initial angular momentum of the system is given by the angular momentum of platform A. The formula for angular momentum is \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.The moment of inertia of platform A, \( I_A = m_A k_A^2 \), where \( m_A = 5 \) kg and \( k_A = 0.175 \) m.\[ I_A = 5 \times (0.175)^2 = 0.153125 \text{ kg}\cdot \text{m}^2 \]The initial angular velocity \( \omega_i \) in radians per second is \( \omega_i = 50 \times \frac{2\pi}{60} = \frac{5\pi}{3} \) rad/s.So, the initial angular momentum \( L_i = I_A \times \omega_i = 0.153125 \times \frac{5\pi}{3} \).
02

Find the Final Moment of Inertia

Final moment of inertia occurs when disk B is against the rim. The moment of inertia of disk B with respect to the axis through its center is \( I_B = \frac{1}{2} m_B r_B^2 \), where \( m_B = 3 \) kg and \( r_B = 0.08 \) m.\[ I_B = \frac{1}{2} \times 3 \times (0.08)^2 = 0.0096 \text{ kg}\cdot \text{m}^2 \]When disk B is against the rim, its distance from the axis is 0.2 m and its moment of inertia relative to the shaft is \( m_B \times 0.2^2 \).Disk B's effective moment of inertia when against the rim is:\[ I_{B,rim} = 3 \times 0.2^2 = 0.12 \text{ kg}\cdot \text{m}^2 \]The total moment of inertia now is:\[ I_{total} = I_A + I_{B,rim} = 0.153125 + 0.12 = 0.273125 \text{ kg}\cdot \text{m}^2 \]
03

Use Conservation of Angular Momentum

Angular momentum is conserved since there is no external torque applied. Thus, \( L_i = L_f \) where \( L_f \) is the final angular momentum.Initially:\[ L_i = I_A \times \omega_i = 0.153125 \times \frac{5\pi}{3} \approx 0.8029 \text{ kg}\cdot \text{m}^2 \cdot \/ \text{s} \]Finally:\[ L_f = I_{total} \times \omega_f \]Setting \( L_i = L_f \), we have:\[ 0.8029 = 0.273125 \times \omega_f \]Solving for \( \omega_f \):\[ \omega_f = \frac{0.8029}{0.273125} \approx 2.940 \/ \text{rad/s} \]
04

Convert Angular Velocity to RPM

To convert the final angular velocity back to rpm for easier interpretation and comparison with the initial condition:\[ \omega_f = 2.940 \/ \text{rad/s} \]Convert rad/s to rpm using:\[ \text{rpm} = \omega_f \times \frac{60}{2\pi} \approx 28.05 \/ \text{rpm} \]The final angular velocity is approximately 28.05 rpm.

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Key Concepts

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

Angular Momentum
Angular momentum is a key concept in rotational dynamics. It is analogous to linear momentum in translational motion but for rotating bodies. The angular momentum of a rotating object depends on two main factors: the moment of inertia (which is the rotational equivalent of mass) and the angular velocity. Mathematically, it is expressed as \( L = I \omega \), where:
  • \( L \) is the angular momentum,
  • \( I \) is the moment of inertia,
  • \( \omega \) is the angular velocity.
In the context of our exercise, the platform initially has angular momentum due to its rotational motion. When the disk B is placed on it and moves towards the rim, it affects the system's total angular momentum. However, since no external torques act on the platform-disk system, the initial and final angular momentum remain equal, showcasing the principle of conservation.
Conservation of Angular Momentum
Conservation of angular momentum is a fundamental principle in physics that states that if no external torque is acting on a system, its total angular momentum remains constant over time. This principle can simplify the analysis of many problems involving rotational motion.Let's get practical: When disk B, initially not rotating, is placed on the spinning platform, it has to speed up to catch up with the platform's motion. As it slides to the rim, it increases the overall moment of inertia of the platform-disk system. Provided there's no slipping or external influence like friction from an outside source, the only force acting is internal, maintaining the system's angular momentum. That's why the initial angular momentum \( L_i = I_A \omega_i \) is equal to the final angular momentum \( L_f = I_{total} \omega_f \). This allows us to solve for the final angular velocity \( \omega_f \).
Angular Velocity
Angular velocity is the rate at which an object rotates around an axis. It is usually measured in radians per second (rad/s) or revolutions per minute (rpm). In our scenario, initially, the platform has an angular velocity of 50 rpm. However, when converted to rad/s (as calculations typically use this unit), it is \( \frac{5\pi}{3} \) rad/s.Understandably, angular velocity helps us predict how quickly the platform spins. When disk B comes to rest against the platform's rim, the system's combined mass moment of inertia alters the angular velocity in a way to conserve angular momentum. We noticed this shift in angular velocity when disk B moved closer to the rim, ultimately lowering the rotational speed as calculated to approximately 28.05 rpm. Calculations involving these changes ensure accuracy in determining the effects of rotational dynamics.

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Most popular questions from this chapter

Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs 12,000 lbs, and the team of oxen generates a constant pull force of 1500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine (a) the velocity of its center G after it has moved 5 ft, (b) the minimum static coefficient of friction that will keep it from slipping.

Knowing that the maximum allowable couple that can be applied to a shaft is 15.5 kip?in., determine the maximum horsepower that can be transmitted by the shaft at (a) 180 rpm, (b) 480 rpm.

A satellite has a total weight (on Earth) of 250 lbs, and each of the solar panels weighs 15 lbs. The body of the satellite has mass moment of inertia about the z-axis of 6 slug-ft \(^{2}\), and the panels can be modeled as flat plates. The satellite spins with a rate of 10 rpm about the z-axis when the solar panels are positioned in the xy plane. Determine the spin rate about z after a motor on the satellite has rotated both panels to be positioned in the yz plane.

Sphere \(A\) of mass \(m_{A}=2 \mathrm{kg}\) and radius \(r=40 \mathrm{mm}\) rolls without slipping with a velocity \(\bar{v}_{1}=2 \mathrm{m} / \mathrm{s}\) on a horizontal surface when it hits squarely a uniform slender bar \(B\) of mass \(m_{B}=0.5 \mathrm{kg}\) and length \(L=100 \mathrm{mm}\) that is standing on end and is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the sphere and the horizontal surface, neglecting friction between the sphere and the bar, and knowing the coefficient of restitution between A and B is 0.1, determine the angular velocities of the sphere and the bar immediately after the impact.

A slender rod of length \(l\) is pivoted about a point \(C\) located at a distance \(b\) from its center \(G .\) It is released from rest in a horizontal position and swings freely. Determine \((a)\) the distance \(b\) for which the angular velocity of the rod as it passes through a vertical position is maximum, \((b)\) the corresponding values of its angular velocity and of the reaction at \(C .\)
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