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Chapter 17: Problem 76

In the gear arrangement shown, gears A and C are attached to rod ABC, that is free to rotate about B, while the inner gear B is fixed. Knowing that the system is at rest, determine the magnitude of the couple M that must be applied to rod ABC, if 2.5 s later the angular velocity of the rod is to be 240 rpm clockwise. Gears A and C weigh 2.5 lb each and may be considered as disks of radius 2 in.; rod ABC weighs 4 lb.

Short Answer

Expert verified
Apply the moment of inertia and angular acceleration to find torque M.

Step by step solution

01

Determine the Moment of Inertia

To find the moment of inertia for the system, consider each component separately. The two gears (A and C) can be treated as disks with a mass of 2.5 lb and radius of 2 inches. Use the formula for the moment of inertia of a disk, \(I = \frac{1}{2} m r^2\). First, convert the weight to mass by dividing by the acceleration due to gravity \(g \approx 32 \text{ ft/s}^2 = 384 \text{ in/s}^2\). The mass of each gear is \(\frac{2.5}{384}\) slugs. Thus, the moment of inertia for each gear (A and C) is \(I = \frac{1}{2} \cdot \frac{2.5}{384} \cdot (2)^2\). The rod ABC, assumed as a rod about the center, has its own inertia which is generally \(\frac{1}{12} m L^2\), where \(L\) is its length. The length isn't given, but all inertias can be simplified and added together.
02

Convert Angles and Angular Velocities

Convert the angular velocity from revolutions per minute (rpm) to radians per second (rad/s): \(240 \text{ rpm} = 240 \times \frac{2\pi}{60} \text{ rad/s}\). Also, convert any necessary linear velocities to angular ones as required by the problem. This step ensures you are using consistent units throughout the calculation.
03

Determine Angular Acceleration

Calculate the angular acceleration \(\alpha\) using the formula \(\alpha = \frac{\Delta\omega}{\Delta t}\). The change in angular velocity \(\Delta\omega\) is from 0 to \(4\pi\) rad/s over a time \(\Delta t = 2.5\) seconds. Thus, \(\alpha = \frac{4\pi}{2.5}\).
04

Apply the Equation of Motion

Use the equation for rotational dynamics, \(M = I\alpha\), where \(M\) is the required moment (torque) to be applied, \(I\) is the total moment of inertia found in Step 1, and \(\alpha\) is the angular acceleration from Step 3. Substitute the calculated values to find \(M\).
05

Calculate Total Moment of Inertia and Solve for M

Calculate the total moment of inertia by adding the individual inertias of the gears and the rod. Use this total in the equation \(M = I\alpha\) to solve for the magnitude \(M\). Ensure units are consistent to find \(M\) in pound-inches or your desired unit.

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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 dealing with rotational motion. It tells us how fast an object spins around a central point or axis. We often measure it in revolutions per minute (rpm) or radians per second (rad/s). To compare or use angular velocity in calculations, like in our gear system exercise, it's crucial to use radians per second.
To convert from rpm to rad/s, multiply the rpm value by \(\frac{2\pi}{60}\), because one revolution is \(2\pi\) radians, and there are 60 seconds in a minute.
In this exercise, the rod's angular velocity is given as 240 rpm, converting this helps in calculating other dynamic properties.
Moment of Inertia
The moment of inertia is like mass in linear motion but for rotation. It measures an object's resistance to changes in its rotational motion. Different shapes and mass distributions have different inertias. For rotational dynamics, knowing the moment of inertia helps in predicting how easily an object will spin given a certain torque.
For typical shapes like disks or rods, formulas can simplify finding this value:
  • For a disk, it's \( I = \frac{1}{2}mr^2 \)
  • For a rod about its center, it's \( I = \frac{1}{12}mL^2 \)
In our problem, we calculate the moment of inertia for both gears as disks and add them, alongside the rod, to find the total inertia for our gear system.
Angular Acceleration
Angular acceleration measures how quickly the angular velocity changes. It represents the torque's effect over time on a rotational system, similar to how linear acceleration relates to force.
The formula for angular acceleration \( \alpha \) is: \[ \alpha = \frac{\Delta\omega}{\Delta t} \]where \( \Delta\omega \) is the change in angular velocity and \( \Delta t \) is the time period over which this change happens.
In the exercise, the rod changes its speed from 0 to its final speed in 2.5 seconds, allowing us to calculate the precise angular acceleration using the change in angular velocity computed from rpm to rad/s.
Rotational Dynamics
Rotational dynamics explains how torques cause objects to rotate, similar to how forces cause linear motion. The central equation, \[ M = I\alpha \] links torque (\( M \)), moment of inertia (\( I \)), and angular acceleration (\( \alpha \)). Here, torque is the rotational equivalent of force, needed to spin an object.
This equation helps solve problems where you know certain conditions and want to find the magnitude of force or torque required, as in the exercise. By finding the moment of inertia and angular acceleration, you can determine the applied moment (torque) necessary to achieve the desired rotational outcome. This foundational principle applies broadly across mechanical systems, like the gears in the problem.

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Most popular questions from this chapter

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.

The \(4-\mathrm{kg}\) rod \(A B\) can slide freely inside the 6 -kg tube. The rod was entirely within the tube \((x=0)\) and released with no initial velocity relative to the tube when the angular velocity of the assembly was 5 rad/s. Neglecting the effect of friction, determine the speed of the rod relative to the tube when \(x=400 \mathrm{mm} .\)

Show that the sum \(\mathbf{H}_{A}\) of the moments about a point \(A\) of the momenta of the particles of a rigid body in plane motion is equal to \(I_{A} \omega,\) where \(\omega\) is the angular velocity of the body at the instant considered and \(I_{A}\) the moment of inertia of the body about \(A,\) if and only if one of the following conditions is satisfied: \((a) A\) is the mass center of the body, (b) \(A\) is the instantaneous center of rotation, (c) the velocity of \(A\) is directed along a line joining point \(A\) and the mass center \(G .\)

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 \(\theta=135^{\circ} .\)

A uniform slender rod \(A B\) has a mass \(m,\) a length \(L,\) and is falling freely with a velocity \(\mathbf{v}_{0}\) when end \(B\) strikes a smooth inclined surface as shown. Assuming that the impact is perfectly elastic, determine the angular velocity of the rod and the velocity of its mass center immediately after the impact.
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