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The propeller of a light plane has a length of \(2.092 \mathrm{~m}\) and a mass of \(17.56 \mathrm{~kg}\). The rotational energy of the propeller is \(422.8 \mathrm{~kJ}\). What is the rotational frequency of the propeller (in rpm)? You can treat the propeller as a thin rod rotating about its center.

Short Answer

Expert verified
Solution: 1. Calculate the moment of inertia: \(I = \frac{1}{12} \times 17.56 \times (2.092)^2\) 2. Rearrange the formula for rotational energy: \(\omega = \sqrt{\frac{2E}{I}}\) 3. Calculate the angular velocity: \(\omega = \sqrt{\frac{2 \times 422800}{I}}\) 4. Convert the angular velocity to rotational frequency: \(n = \frac{\omega}{2\pi} \times 60\)

Step by step solution

01

Find the moment of inertia of the propeller

To find the moment of inertia of the propeller, we can treat it as a thin rod rotating about its center. The moment of inertia for a thin rod is given by the formula \(I = \frac{1}{12}mL^2\), where m is the mass of the rod and L is its length. Given, mass \(m = 17.56 \mathrm{~kg}\) and length \(L = 2.092 \mathrm{~m}\), we can find the moment of inertia: \(I = \frac{1}{12} \times 17.56 \times (2.092)^2\)
02

Use the formula for rotational energy

The rotational energy of the propeller is given as \(E = 422.8 \mathrm{~kJ} = 422800 \mathrm{~J}\). The formula for rotational energy is \(E = \frac{1}{2}I\omega^2\), where \(\omega\) is the angular velocity and \(E\) is the energy. We can rearrange this formula to find the angular velocity: \(\omega^2 = \frac{2E}{I} \implies \omega = \sqrt{\frac{2E}{I}}\)
03

Calculate the angular velocity

Now, we can use the rotational energy and the moment of inertia to calculate the angular velocity: \(\omega = \sqrt{\frac{2 \times 422800}{I}}\) After finding the moment of inertia in step 1, substitute its value in this formula.
04

Convert the angular velocity to rotational frequency

The angular velocity is given in radians per second, but we need the rotational frequency in revolutions per minute. We can use the following conversion: \(n = \frac{\omega}{2\pi} \times 60\) Where \(n\) is the rotational frequency in revolutions per minute (rpm), \(\omega\) is the angular velocity in radians per second, and the factor \(60\) is used to convert from seconds to minutes. Using the angular velocity calculated in step 3: \(n = \frac{\omega}{2\pi} \times 60\) This will give us the rotational frequency of the propeller in rpm.

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Key Concepts

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

Moment of Inertia
The moment of inertia, often represented as 'I', is a measure of an object’s resistance to changes in its rotation rate. It plays a critical role in rotational dynamics, just as mass does in linear motion. For any given object, its moment of inertia depends on the distribution of its mass relative to the axis of rotation.

To find the moment of inertia for different shapes, various formulas are used. For a thin rod rotating about its center, the formula is: \( I = \frac{1}{12}mL^2 \). Here, 'm' signifies the mass of the rod, and 'L' is its length. The moment of inertia is directly proportional to both the mass of the object and the square of its rotational length. This means that the further the mass is distributed from the rotational axis, the greater the moment of inertia, and consequently, the more force it requires to change its rotational speed.

In the case of the propeller exercise, the mass and length of the propeller were given, enabling the calculation of its moment of inertia. This value serves as a foundation to assess the rotational energy and frequency, as it quantifies how much kinetic energy can be stored in the propeller's rotation given its mass distribution.
Rotational Energy
Rotational energy, also known as rotational kinetic energy, is the energy possessed by a rotating object due to its motion. It is analogous to linear kinetic energy but for rotating systems. The formula to calculate rotational energy is \( E = \frac{1}{2}I\omega^2 \), where 'E' represents the rotational energy, 'I' is the moment of inertia, and 'ω' (omega) is the angular velocity.

This formula shows how rotational energy is influenced by both the moment of inertia and the square of the angular velocity. Thus, an object having a high moment of inertia or rotating at a high speed will have substantial rotational energy. For instance, in the exercise with the airplane propeller, knowing the rotational energy and the moment of inertia allows us to solve for the angular velocity. This is vital because it indicates how the propeller's mass distribution and speed contribute to the amount of stored energy. Since the propeller's rotational energy was provided, it could be used to find the angular velocity, crucial for obtaining the rotational frequency.
Angular Velocity
Angular velocity, denoted by 'ω', measures the rate of rotation of an object. It is defined as the angle rotated per unit of time and is typically measured in radians per second (rad/s). In the context of the propeller problem, angular velocity is essential because it determines the speed at which the propeller is rotating.

Angular velocity is related to the kinetic energy of rotation. As the formula \( \omega = \sqrt{\frac{2E}{I}} \) indicates, angular velocity can be calculated if the rotational energy and the moment of inertia are known. Once 'ω' is found, it can be converted into rotational frequency, which is commonly measured in revolutions per minute (rpm). The formula for this conversion is \( n = \frac{\omega}{2\pi} \times 60 \).

The practical application of understanding angular velocity goes beyond exercises. It’s crucial in engineering, where the operational speeds of components are significant for both functionality and safety. For instance, in the propeller exercise, determining the angular velocity would ensure the propeller operates within safe limits while maintaining the effectiveness of the airplane's propulsion system.

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Most popular questions from this chapter

The flywheel of an old steam engine is a solid homogeneous metal disk of mass \(M=120 . \mathrm{kg}\) and radius \(R=80.0 \mathrm{~cm} .\) The engine rotates the wheel at \(500 .\) rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force \(F=100 .\) N. If the coefficient of kinetic friction between the pad and the flywheel is \(\mu_{\mathrm{k}}=0.200\), how many revolutions does the flywheel make before coming to rest? How long does it take for the flywheel to come to rest? Calculate the work done by the torque during this time.

A basketball of mass \(610 \mathrm{~g}\) and circumference \(76 \mathrm{~cm}\) is rolling without slipping across a gymnasium floor. Treating the ball as a hollow sphere, what fraction of its total kinetic energy is associated with its rotational motion? a) 0.14 b) 0.19 c) 0.29 d) 0.40 e) 0.67

It is sometimes said that if the entire population of China stood on chairs and jumped off simultaneously, it would alter the rotation of the Earth. Fortunately, physics gives us the tools to investigate such speculations. a) Calculate the moment of inertia of the Earth about its axis. For simplicity, treat the Earth as a uniform sphere of mass \(m_{\mathrm{E}}=5.977 \cdot 10^{24} \mathrm{~kg}\) and radius \(6371 \mathrm{~km}\). b) Calculate an upper limit for the contribution by the population of China to the Earth's moment of inertia, by assuming that the whole group is at the Equator. Take the population of China to be 1.30 billion people, of average mass \(70.0 \mathrm{~kg}\) c) Calculate the change in the contribution in part (b) associated with a \(1.00-\mathrm{m}\) simultaneous change in the radial position of the entire group. d) Determine the fractional change in the length of the day the change in part (c) would produce.

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 . \mathrm{kg}\) rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An \(80.0-\mathrm{kg}\) man standing at the very center of the platform starts walking (at \(t=0\) ) radially outward at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\) with respect to the platform. Approximating the man by a vertical cylinder of radius \(R_{\mathrm{m}}=0.200 \mathrm{~m},\) determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

A uniform solid sphere of radius \(R\) and mass \(M\) sits on a horizontal tabletop. A horizontally directed impulse with magnitude \(J\) is delivered to a spot on the ball a vertical distance \(h\) above the tabletop. a) Determine the angular and translational velocity of the sphere just after the impulse is delivered. b) Determine the distance \(h_{0}\) at which the delivered impulse causes the ball to immediately roll without slipping.

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