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Chapter 10: Problem 35

Determine the moment of inertia for three children weighing \(60.0 \mathrm{lb}, 45.0 \mathrm{lb}\) and \(80.0 \mathrm{lb}\) sitting at different points on the edge of a rotating merry-go-round, which has a radius of \(12.0 \mathrm{ft}\).

Short Answer

Expert verified
Answer: The moment of inertia for the three children sitting on the edge of the merry-go-round is approximately \(1121.993 \mathrm{kg \cdot m^2}\).

Step by step solution

01

Convert weights to mass

First, we need to convert the weights of the children from pounds (lb) to mass in kilograms (kg). To do this, we can use the following conversion factor: 1 lb = 0.453592 kg. Child 1: \(60.0 \mathrm{lb} \times 0.453592 \frac{\mathrm{kg}}{\mathrm{lb}} = 27.216 \mathrm{kg}\) Child 2: \(45.0 \mathrm{lb} \times 0.453592 \frac{\mathrm{kg}}{\mathrm{lb}} = 20.412 \mathrm{kg}\) Child 3: \(80.0 \mathrm{lb} \times 0.453592 \frac{\mathrm{kg}}{\mathrm{lb}} = 36.288 \mathrm{kg}\)
02

Convert the radius to meters

Next, we need to convert the radius of the merry-go-round from feet (ft) to meters (m). To do this, we can use the following conversion factor: 1 ft = 0.3048 m. Radius: \(12.0 \mathrm{ft} \times 0.3048 \frac{\mathrm{m}}{\mathrm{ft}} = 3.6576 \mathrm{m}\)
03

Calculate the moment of inertia for each child

Now, we can calculate the moment of inertia for each child using the formula \(I = m \times r^2\). Child 1: \(I_1 = 27.216 \mathrm{kg} \times (3.6576 \mathrm{m})^2 = 364.172 \mathrm{kg\cdot m^2}\) Child 2: \(I_2 = 20.412 \mathrm{kg} \times (3.6576 \mathrm{m})^2 = 273.018 \mathrm{kg\cdot m^2}\) Child 3: \(I_3 = 36.288 \mathrm{kg} \times (3.6576 \mathrm{m})^2 = 484.803 \mathrm{kg\cdot m^2}\)
04

Calculate the total moment of inertia

Finally, we can find the total moment of inertia by adding the moments of inertia for each child. Total moment of inertia: \(I_{total} = I_1 + I_2 + I_3 = 364.172 + 273.018 + 484.803 = 1121.993 \mathrm{kg\cdot m^2}\) So, the moment of inertia for the three children sitting on the edge of the merry-go-round is approximately \(1121.993 \mathrm{kg \cdot m^2}\).

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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 is a property of any object that describes how difficult it is to change its rotational motion about an axis. Think of it as the rotational equivalent of mass in linear motion. For a point mass, the moment of inertia is calculated by the product of the mass and the square of the distance to the rotation axis, represented by the formula:
\( I = m \times r^2 \)
In the context of our merry-go-round problem, each child represents a point mass located at a distance from the axis of rotation (the center). The further the child is from the center, the higher the moment of inertia. It's essential to understand that moment of inertia is additive for multiple point masses, which means you can simply add up the moment of inertia of each child to get the total moment of inertia of the system.
Physics Problem Solving
Problem solving in physics often follows a systematic approach that involves identifying the problem, gathering information, forming a plan, applying physics concepts and equations, and finally, solving the problem. In our example, we first identified the aim, which is to calculate the moment of inertia for children on a merry-go-round. We gathered information on the weights of the children and the radius of the merry-go-round. Then we formed a plan by determining the steps needed, including unit conversions and the application of the moment of inertia formula. Understanding the underlying concepts like moment of inertia and applying it alongside mathematical operations is key to solving physics problems effectively.
Unit Conversion
Unit conversion is crucial in physics to ensure all quantities are expressed in compatible units before performing calculations. In our example, we converted weights from pounds to kilograms and the radius from feet to meters, using the conversion factors \(1 lb = 0.453592 kg\) and \(1 ft = 0.3048 m\) respectively. Mastering unit conversions is often the first step in solving a physics problem since the correctness of the final result heavily depends on using the correct units throughout the calculation. This requires familiarity with the metric system, which is the standard system of measurement in physics.

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Most popular questions from this chapter

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 .\) kg rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An 80.0 -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?

To turn a motorcycle to the right, you do not turn the handlebars to the right, but instead slightly to the left. Explain, as precisely as you can, how this counter-steering turns the motorcycle in the desired direction. (Hint: The wheels of a motorcycle in motion have a great deal of angular momentum.)

In experiments at the Princeton Plasma Physics Laboratory, a plasma of hydrogen atoms is heated to over 500 million degrees Celsius (about 25 times hotter than the center of the Sun) and confined for tens of milliseconds by powerful magnetic fields \((100,000\) times greater than the Earth's magnetic field). For each experimental run, a huge amount of energy is required over a fraction of a second, which translates into a power requirement that would cause a blackout if electricity from the normal grid were to be used to power the experiment. Instead, kinetic energy is stored in a colossal flywheel, which is a spinning solid cylinder with a radius of \(3.00 \mathrm{~m}\) and mass of \(1.18 \cdot 10^{6} \mathrm{~kg}\). Electrical energy from the power grid starts the flywheel spinning, and it takes 10.0 min to reach an angular speed of \(1.95 \mathrm{rad} / \mathrm{s}\). Once the flywheel reaches this angular speed, all of its energy can be drawn off very quickly to power an experimental run. What is the mechanical energy stored in the flywheel when it spins at \(1.95 \mathrm{rad} / \mathrm{s}\) ? What is the average torque required to accelerate the flywheel from rest to \(1.95 \mathrm{rad} / \mathrm{s}\) in \(10.0 \mathrm{~min} ?\)

An ice skater spins with her arms extended and then pulls her arms in and spins faster. Which statement is correct? a) Her kinetic energy of rotation does not change because, by conservation of angular momentum, the fraction by which her angular velocity increases is the same as the fraction by which her rotational inertia decreases. b) Her kinetic energy of rotation increases because of the work she does to pull her arms in. c) Her kinetic energy of rotation decreases because of the decrease in her rotational inertia; she loses energy because she gradually gets tired.

A uniform rod of mass \(M=250.0 \mathrm{~g}\) and length \(L=50.0 \mathrm{~cm}\) stands vertically on a horizontal table. It is released from rest to fall. a) What forces are acting on the rod? b) Calculate the angular speed of the rod, the vertical acceleration of the moving end of the rod, and the normal force exerted by the table on the rod as it makes an angle \(\theta=45.0^{\circ}\) with respect to the vertical. c) If the rod falls onto the table without slipping, find the linear acceleration of the end point of the rod when it hits the table and compare it with \(g\).
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