/*! 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 36 An airplane propeller is 2.08 \(... [FREE SOLUTION] | 91Ó°ÊÓ

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An airplane propeller is 2.08 \(\mathrm{m}\) in length (from tip to tip) with mass 117 \(\mathrm{kg}\) and is rotating at 2400 \(\mathrm{rpm}(\mathrm{rev} / \mathrm{min})\) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0\(\%\) of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

Short Answer

Expert verified
(a) Rotational kinetic energy: approximately 11510 J; (b) New angular speed: approximately 2771 rpm.

Step by step solution

01

Determine the Moment of Inertia

The moment of inertia of a slender rod rotating about its center is given by the equation \( I = \frac{1}{12} m L^2 \), where \( m \) is the mass and \( L \) is the length. For our propeller, \( m = 117 \, \mathrm{kg} \) and \( L = 2.08 \, \mathrm{m} \). Thus, \( I = \frac{1}{12} \times 117 \times (2.08)^2 \).
02

Calculate Rotational Kinetic Energy

The rotational kinetic energy is calculated using \( KE = \frac{1}{2} I \omega^2 \), where \( \omega \) is the angular velocity in rad/s. First, convert the angular speed from rpm to rad/s: \( 2400 \, \mathrm{rpm} \times \frac{2\pi}{60} = \omega \). Substitute \( I \) and the converted \( \omega \) into the kinetic energy formula to find \( KE \).
03

Calculate new Mass and Moment of Inertia

Reduce the mass to 75% of its original: \( m_{new} = 0.75 \times 117 \, \mathrm{kg} \). Recalculate the moment of inertia \( I_{new} \) using \( I_{new} = \frac{1}{12} m_{new} L^2 \).
04

Solve for New Angular Speed

To maintain the same kinetic energy, we use \( KE = \frac{1}{2} I_{new} \omega_{new}^2 \). Since \( KE \) is constant, equate \( \frac{1}{2} I \omega^2 = \frac{1}{2} I_{new} \omega_{new}^2 \) and solve for \( \omega_{new} \). Convert \( \omega_{new} \) from rad/s back to rpm to find the required angular speed.

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Key Concepts

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

Understanding the Moment of Inertia
The moment of inertia is a crucial concept in rotational kinetics that describes how difficult it is to change an object's rotational motion. For a slender rod, such as an airplane propeller rotating about its center, the moment of inertia, denoted as \( I \), is calculated with the formula: \[ I = \frac{1}{12} m L^2 \]where \( m \) represents the mass, and \( L \) is the length from tip to tip.The moment of inertia depends fundamentally on the mass distribution relative to the axis of rotation. - Heavier or longer objects have a higher moment of inertia, making them harder to spin.- Mass located further from the pivot contributes more significantly to the moment of inertia.In the case of the airplane propeller, the given mass \( 117 \; \, \text{kg} \) and length \( 2.08 \; \text{m} \) allow us to compute the inertia needed for our initial calculations.
Explaining Angular Velocity
Angular velocity is akin to linear velocity but for rotational or circular motion. It measures how fast an object is spinning around a center or axis and is usually expressed in radians per second (rad/s). To convert from revolutions per minute (rpm) to radians per second, we use:\[ \omega = \text{rpm} \times \frac{2\pi}{60} \]For the airplane propeller spinning at 2400 rpm, performing this conversion yields the angular velocity required to compute the rotational kinetic energy.Angular velocity is vital because it helps in understanding the dynamics of rotating systems. By knowing how fast an object turns, we can determine other essential parameters such as kinetic energy or apply it in conservation equations. - More angular velocity means faster rotation.- Different applications may need specific angular velocities for optimal performance.
The Role of Conservation of Energy
The conservation of energy principles tell us that energy cannot be created or destroyed; it can only change forms. In the context of rotational dynamics, we apply this concept to ensure that the total energy before and after any changes remains constant. The rotational kinetic energy \( KE \) can be described as:\[ KE = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.When reducing the mass of a propeller to 75% of its original mass, we want to maintain the same kinetic energy. This situation requires an adjustment to the angular velocity to compensate for the reduced moment of inertia:- With less mass, the propeller's resistance to rotational motion decreases.- To keep the kinetic energy constant, the velocity of rotation has to change accordingly.
Impact of Mass Reduction
The exercise involves reducing the propeller's mass to 75% of its initial value to meet specific weight requirements. This reduction affects the moment of inertia, calculated with the updated mass value:\[ m_{new} = 0.75 \times m_{original} \]The new moment of inertia \( I_{new} \) can be determined similarly: \[ I_{new} = \frac{1}{12} m_{new} L^2 \]With the mass reduction:- There's a lower resistance to rotational change which can influence how easily an object spins.- The aircraft will benefit from reduced overall weight, which can enhance fuel efficiency.To preserve the kinetic energy, the angular velocity must be recalibrated so that \( KE \) remains unchanged, ensuring the propeller functions effectively at its new mass.
Converting Angular Speed Across Units
Angular speed conversion is key when transitioning between different angular units, like converting from rad/s to rpm or vice versa. In problems involving heavy machinery or aircraft, such conversions are routine. For instance, the formula used in conversion is:\[ \omega_{new \text{ rpm}} = \omega_{new \text{ rad/s}} \times \frac{60}{2\pi} \]After calculating the new angular speed to maintain the same kinetic energy despite mass reduction, it's useful to convert it back for practical applications.Understanding these conversions allows engineers and designers to ensure systems operate at required speeds:- It facilitates communication between units commonly used in engineering specifications.- By translating speeds accurately into other formats, optimal performance standards are assured across different system aspects.

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Most popular questions from this chapter

At \(t=0\) a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 rad/s \(^{2}\) until a circuit breaker trips at \(t=2.00\) s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by moans of a V-belt. A circular saw blade of diameter 0.208 \(\mathrm{m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't to its teeth.

9.75\(\cdots .\) It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) ) in the shape of a 10.0 -cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

A turntable rotates with a constant 2.25 \(\mathrm{rad} / \mathrm{s}^{2}\) angular acceleration. After 4.00 s it has rotated through an angle of 60.0 rad. What was the angular velocity of the wheel at the beginning of the 4.00 -s interval?

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