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A uniform 144 -lb cube is attached to a uniform \(136-\) -lb circular shaft as shown, and a couple \(\mathrm{M}\) with a constant magnitude is applied to the shaft when the system is at rest. Knowing that \(r=4\) in. \(L=12 \mathrm{in.}\), and the angular velocity of the system is \(960 \mathrm{rpm}\) after \(4 \mathrm{s}\), determine the magnitude of the couple \(\mathrm{M}\).

Short Answer

Expert verified
The magnitude of the couple \(M\) is approximately 4086 lb-in.

Step by step solution

01

Convert Units

First, convert all given measurements to consistent units. We have the mass of the cube as 144 lb and the mass of the shaft as 136 lb. Convert pounds to slugs by dividing by the acceleration due to gravity (32.2 ft/s²). Hence, the masses in slugs are \( \frac{144}{32.2} \approx 4.47 \text{ slugs} \) for the cube and \( \frac{136}{32.2} \approx 4.22 \text{ slugs} \) for the shaft. Also, convert the radius and length of the shaft from inches to feet: \( r = \frac{4}{12} = \frac{1}{3} \text{ ft} \) and \( L = \frac{12}{12} = 1 \text{ ft} \).
02

Calculate Moment of Inertia

Calculate the moment of inertia (I) of the system. For the cube rotating around its center, the inertial moment is \( I_\text{cube} = \frac{1}{6} \times m_\text{cube} \times (\frac{L}{2})^2 = \frac{1}{6} \times 4.47 \times (0.5)^2 \approx 0.186 \text{ slug ft}^2 \). For the shaft, it's a cylinder rotating about its center axis: \( I_\text{shaft} = \frac{1}{2} \times m_\text{shaft} \times r^2 = \frac{1}{2} \times 4.22 \times (\frac{1}{3})^2 \approx 0.235 \text{ slug ft}^2 \). The total \( I = I_\text{cube} + I_\text{shaft} = 0.186 + 0.235 = 0.421 \text{ slug ft}^2 \).
03

Convert Angular Velocity

Convert the angular velocity from rpm to rad/s. The given angular speed is 960 rpm. To convert to rad/s: \( \omega = 960 \times \frac{2\pi}{60} \approx 100.53 \text{ rad/s} \).
04

Determine Angular Acceleration

Find angular acceleration (\(\alpha\)) using \(\alpha = \frac{\Delta \omega}{\Delta t}\). We start from rest, so \( \Delta \omega = 100.53 - 0 = 100.53 \text{ rad/s} \) in 4 seconds \( \Delta t = 4 \text{ s} \), thus \( \alpha = \frac{100.53}{4} \approx 25.13 \text{ rad/s}^2 \).
05

Use Newton's Second Law for Rotation

Apply Newton's second law for rotation \(M = I \cdot \alpha\). Substitute the total moment of inertia and angular acceleration: \[M = 0.421 \times 25.13 \approx 10.57 \text{ slug ft}^2 / s^2\]. Convert this to pound-inches by multiplying by 32.2 and converting feet to inches \( M \approx 10.57 \times 32.2 \times 12 \approx 4086 \text{ lb-in} \).
06

Conclusion

The calculated magnitude of the couple M is approximately 4086 pound-inches, considering the given data and the rotational motion around the axis.

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Key Concepts

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

Moment of Inertia
Understanding moment of inertia is crucial in rotational dynamics as it plays a role similar to mass in linear motion. Moment of inertia is essentially a measure of how much torque is required for a desired angular acceleration about an axis of rotation. It depends not only on the mass but also on how that mass is distributed relative to the axis.

Take, for instance, a cube rotating around an axis through its center. The formula for the moment of inertia of a cube is \[ I_\text{cube} = \frac{1}{6} \times m \times (\frac{L}{2})^2 \]where \( m \) is the mass of the cube, and \( L \) is the length of its sides. For a rotating cylinder, like the shaft in the exercise, the moment of inertia is determined by \[ I_\text{shaft} = \frac{1}{2} \times m \times r^2 \]where \( r \) is the radius.

Calculating the total moment of inertia involves adding up the individual moments of inertia of all components of a system, giving insights into how the system will react under applied torques.
Angular Velocity
Angular velocity is the rate at which an object rotates or spins around an axis. It's a vector quantity with both magnitude and direction, typically expressed in radians per second (rad/s).At a conversion checkpoint, you might need to change units from revolutions per minute (rpm) to rad/s, especially when applying formulas in physics. The conversion is straightforward using:

\[ \omega = \text{rpm} \times \frac{2\pi}{60} \]

For instance, converting 960 rpm involves plugging into the conversion formula to get radians per second. This conversion is critical because physics equations in rotational dynamics use SI units (radians and seconds) for consistency and accuracy.With this computation, students can then confidently engage with formulas to find angular acceleration and apply the correct units, paving the way for deeper understanding of rotational motion.
Newton's Second Law for Rotation
Newton's second law for rotation defines how a net torque impacts an object's rotational motion, analogous to how force affects linear movement in Newton's first law. The law is encapsulated in the formula \[ M = I \cdot \alpha \]where \( M \) is the net torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.

This formula allows us to connect the moment of inertia with the resulting angular acceleration, demonstrating how external torques affect an object's spin. In the exercise, calculating torque required the understanding of angular acceleration drawn from the change in angular velocity over time.

By calculating the moment of inertia for each part and applying it with angular acceleration, the Newton's second law for rotation helps in determining the magnitude of the necessary applied torque to achieve a specific rotational motion in real-world applications.

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Most popular questions from this chapter

A \(1.6-\mathrm{kg}\) tube \(A B\) can slide freely on rod \(D E,\) which in turn can rotate freely in a horizontal plane. Initially the assembly is rotating with an angular velocity of magnitude \(\omega=5\) rad/s and the tube is held in position by a cord. The moment of inertia of the rod and bracket about the vertical axis of rotation is \(0.30 \mathrm{kg} \cdot \mathrm{m}^{2}\) and the centroidal moment of inertia of the tube about a vertical axis is \(0.0025 \mathrm{kg} \cdot \mathrm{m}^{2}\). If the cord suddenly breaks, determine \((a)\) the angular velocity of the assembly after the tube has moved to end \(E,(b)\) the energy lost during the plastic impact at \(E .\)

A uniform slender bar of length \(L=200 \mathrm{mm}\) and mass \(m=0.5 \mathrm{kg}\) is supported by a frictionless horizontal table. Initially the bar is spinning about its mass center \(G\) with a constant angular speed \(\omega_{1}=6 \mathrm{rad} / \mathrm{s}\). Suddenly latch \(D\) is moved to the right and is struck by end \(A\) of the bar. Knowing that the coefficient of restitution between \(A\) and \(D\) is \(e=0.6\), determine the angular velocity of the bar and the velocity of its mass center immediately after the impact.

The rotor of an electric motor has a mass of 25 kg, and it is observed that 4.2 min is required for the rotor to coast to rest from an angular velocity of 3600 rpm. Knowing that kinetic friction produces a couple of magnitude 1.2 N?m, determine the centroidal radius of gyration for the rotor.

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.

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.

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