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The propeller of a light plane has a length of \(1.812 \mathrm{~m}\) and rotates at \(2160 .\) rpm. The rotational kinetic energy of the propeller is \(124.3 \mathrm{~kJ}\). What is the mass of the propeller? You can treat the propeller as a thin rod rotating about its center.

Short Answer

Expert verified
Answer: The mass of the propeller is approximately 7.173 kg.

Step by step solution

01

Write down the rotational kinetic energy formula

The rotational kinetic energy KE_rot is given by the formula: \(\text{KE}_\text{rot} = \frac{1}{2} I \omega^2\) where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
02

Convert angular velocity from RPM to rad/s

The rotational speed is given in revolutions per minute (RPM). To convert this value to radians per second (rad/s), first divide by 60 to get revolutions per second (RPS), then multiply by \(2\pi\) to get rad/s: \(\omega = 2160 \mathrm{~RPM} \times \frac{1}{60} \times 2\pi \mathrm{~rad/s} \approx 226.193 \mathrm{~rad/s}\)
03

Find the moment of inertia for the thin rod

The moment of inertia for a thin rod rotating about its center is given by the formula: \(I = \frac{1}{12} m L^2\) where \(m\) is the mass of the rod and \(L\) is its length.
04

Calculate the moment of inertia using the kinetic energy formula

Now we have the rotational kinetic energy and the angular velocity. We can plug the values into the kinetic energy formula and solve for the moment of inertia: \(124.3 \mathrm{~kJ} = \frac{1}{2} I \times (226.193 \mathrm{~rad/s})^2\) \(I \approx 2.413 \mathrm{~kg \cdot m^2}\)
05

Solve for the mass of the propeller

Now we know the moment of inertia and length of the rod. Substitute these values into the moment of inertia formula, and solve for the mass of the propeller: \(2.413 \mathrm{~kg \cdot m^2} = \frac{1}{12} m \times (1.812 \mathrm{~m})^2\) \(m \approx 7.173 \mathrm{~kg}\) Therefore, the mass of the propeller is approximately \(7.173\) kg.

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Key Concepts

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

Moment of Inertia
Understanding the moment of inertia is crucial for solving physics problems involving rotational motion. It's analogous to mass in linear dynamics and represents the distribution of mass in an object as it spins or rotates. Think of it as the 'rotational mass'. The higher the moment of inertia, the harder it is to change the object's rotational speed.

For different shapes and rotation axes, the moment of inertia has distinct formulas. For a thin rod rotating around its center, as in our propeller example, it is calculated using \(I = \frac{1}{12} m L^2\), where \(m\) is the mass and \(L\) is the length of the rod. This formula assumes that the mass is uniformly distributed along the length of the rod.
Angular Velocity
Angular velocity determines the rate at which an object rotates, expressed as the angle turned per unit of time. It's the rotational equivalent of linear velocity. Commonly represented by \(\omega\), it's measured in radians per second (rad/s) in the International System of Units (SI).

To find angular velocity from revolutions per minute (RPM), as we've done in the propeller problem, you convert RPM to revolutions per second (RPS) by dividing by 60, and then to rad/s by multiplying by \(2\pi\). Understanding this conversion is essential, as many real-life problems, like the spinning of a propeller or wheels, are often initially given in RPM.
Physics Problem Solving
Effective problem solving in physics is about understanding concepts and methodically applying formulas. In the case of our rotating propeller, we've followed a step-by-step approach that involves identifying known quantities, converting units when necessary, applying the appropriate formulas, and algebraically solving for the unknown.

As an important tip, always check if the units you're using are consistent. For example, angular velocity should be in rad/s when using the kinetic energy formula, and the moment of inertia must be in SI units (\(kg\cdot m^2\)) to ensure the calculated kinetic energy is in joules.
Rotational Dynamics
Rotational dynamics involves the forces and torques that cause changes in rotational motion. The principles of rotational dynamics allow us to understand how torques affect the rotational motion, given the object's moment of inertia. It encompasses concepts like torque, angular momentum, and rotational kinetic energy, which in the case of our propeller, is given by \(\text{KE}_\text{rot} = \frac{1}{2} I \omega^2\).

The concept of rotational kinetic energy is particularly important. It represents the energy due to the rotation of an object and is directly proportional to both its moment of inertia and the square of its angular velocity. This concept connects rotational motion to energy, a fundamental aspect of various physical applications, from engineering to astrophysics.

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Most popular questions from this chapter

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.

A golf ball with mass \(45.90 \mathrm{~g}\) and diameter \(42.60 \mathrm{~mm}\) is struck such that it moves with a speed of \(51.85 \mathrm{~m} / \mathrm{s}\) and rotates with a frequency of \(2857 \mathrm{rpm} .\) What is the kinetic energy of the golf ball?

The turbine and associated rotating parts of a jet engine have a total moment of inertia of \(25.0 \mathrm{~kg} \mathrm{~m}^{2} .\) The turbine is accelerated uniformly from rest to an angular speed of \(150 . \mathrm{rad} / \mathrm{s}\) in a time of \(25.0 \mathrm{~s}\). Find a) the angular acceleration, b) the net torque required, c) the angle turned through in \(25.0 \mathrm{~s}\) d) the work done by the net torque, and e) the kinetic energy of the turbine at the end of the \(25.0 \mathrm{~s}\).

A string is wrapped many times around a pulley and is connected to a block of mass \(m_{\mathrm{b}}=4.701 \mathrm{~kg},\) which is hanging vertically. The pulley consists of a wheel of radius \(47.49 \mathrm{~cm},\) with spokes that have negligible mass. The block accelerates downward at \(4.330 \mathrm{~m} / \mathrm{s}^{2}\) What is the mass of the pulley, \(m_{\mathrm{p}}\) ?

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.

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