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Chapter 9: Problem 38

An airplane propeller is \(2.08 \mathrm{~m}\) in length (from tip to tip) with mass \(117 \mathrm{~kg}\) and is rotating at 2400 rpm (rev/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
The rotational kinetic energy of the propeller is roughly about 13596 J. The new angular speed of the propeller, when its mass is reduced to 75% but keeping the same kinetic energy, is approximately 2942 rpm.

Step by step solution

01

Calculate the angular speed in rad/s

Convert the angular speed from rpm to rad/s using the formula: \( \omega = rpm * 2 * \pi / 60 \). Replace rpm with 2400.
02

Calculate the moment of inertia

Calculate the moment of inertia of the propeller using the formula: \( I = 1/12 * m * L^2 \). Replace \( m \) with 117 kg, and \( L \) with 2.08m.
03

Calculate the rotational kinetic energy

Calculate the rotational kinetic energy using the formula: \( KE = 0.5 * I * \omega^2 \). Plug in the values of \( I \) and \( \omega \) obtained from the previous steps.
04

Calculate new moment of inertia

If the propeller's mass is reduced to 75% of its original mass but maintain the same kinetic energy, calculate the new moment of inertia using the same formula. Replace \( m \) with 75% of 117 kg.
05

Calculate new angular speed

Set the new kinetic energy equivalent to the old one: \( 0.5 * I_old * \omega_old^2 = 0.5 * I_new * \omega_new^2 \), get \( \omega_new = \sqrt{I_old/I_new} * \omega_old \). Calculate the new angular speed in rad/s. Then convert it back to rpm using the formula: \( rpm = \omega * 60 / (2 * \pi) \).

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Key Concepts

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

Angular Speed
Angular speed is a measure of how quickly an object rotates around a center point or axis. Imagine spinning a basketball on your finger: the faster it spins, the higher its angular speed. Angular speed is typically measured in radians per second (rad/s) or revolutions per minute (rpm).
To convert from rpm to rad/s, you use the formula:
  • \( \omega = \text{rpm} \times \frac{2 \pi}{60} \).
This formula helps express rotational speed in terms of radians, which is useful in physics because radians are a natural measure of angle in the circles. For example, in our exercise, the airplane propeller spins at 2400 rpm. Using the conversion formula, its angular speed in rad/s can be calculated, which provides a necessary input for further calculations.
Moment of Inertia
Moment of inertia is like mass for rotation. It's how much resistance an object has to changing its rotation. Think of it as how hard it is to spin an object because of its mass distribution. For simple shapes like rods, the calculation is straightforward:
  • The formula is: \( I = \frac{1}{12} m L^2 \).
Here, \( m \) is the mass and \( L \) is the length of the object. Applying it to our propeller example, the original 117 kg and a length of 2.08 meters. You plug these numbers into the formula to find its moment of inertia. If you reduce the mass, as required in the exercise's second question, you simply adjust the mass value in the same equation. This provides the new moment of inertia, an important step in finding how the rotation speed needs to change.
Kinetic Energy Calculation
Rotational kinetic energy is the energy an object possesses due to its rotation. It's akin to the kinetic energy of a car moving down the road, except this energy is about rotation rather than linear movement. The formula to calculate rotational kinetic energy is:
  • \( KE = 0.5 \times I \times \omega^2 \).
Where \( KE \) stands for kinetic energy, \( I \) is the moment of inertia previously calculated, and \( \omega \) is the angular speed. In the exercise, we calculated the propeller's initial rotational kinetic energy using its moment of inertia and angular speed. Then, if the propeller's mass changes, we must adjust either the moment of inertia or the angular speed to maintain the same kinetic energy. By rearranging and solving our formulas, we can find the new angular speed needed to keep the same energy after the mass reduction.

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Most popular questions from this chapter

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

An advertisement claims that a centrifuge takes up only \(0.127 \mathrm{~m}\) of bench space but can produce a radial acceleration of \(3000 g\) at 5000 rev \(/\) min. Calculate the required radius of the centrifuge. Is the claim realistic?

A disk of radius \(25.0 \mathrm{~cm}\) is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. \(\mathbf{P 9 . 6 1}\) ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is \(1.80 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find \(A\). (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of \(15.0 \mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches \(15.0 \mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .)\)

A pulley on a frictionless axle has the shape of a uniform solid disk of mass \(2.50 \mathrm{~kg}\) and radius \(20.0 \mathrm{~cm}\). A \(1.50 \mathrm{~kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has \(4.50 \mathrm{~J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. \(\mathbf{P 9 . 6 4}\) ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is \(3.71 \mathrm{~m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to \(15.0 \mathrm{~kg}\) and allowed to fall through \(5.00 \mathrm{~m},\) it gives \(250.0 \mathrm{~J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0 \mathrm{~kg}\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the \(15.0 \mathrm{~kg}\) mass be moving on Mars just as the drum gained \(250.0 \mathrm{~J}\) of kinetic energy?
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