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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

The turbine and associated rotating parts of a jet engine have a total moment of inertia of \(25 \mathrm{~kg} \mathrm{~m}^{2}\). The turbine is accelerated uniformly from rest to an angular speed of \(150 \mathrm{rad} / \mathrm{s}\) in a time of \(25 \mathrm{~s}\). Find a) the angular acceleration, b) the net torque required, c) the angle turned through in \(25 \mathrm{~s}\) d) the work done by the net torque, and e) the kinetic energy of the turbine at the end of the \(25 \mathrm{~s}\).

A projectile of mass \(m\) is launched from the origin at speed \(v_{0}\) at angle \(\theta_{0}\) above the horizontal. Air resistance is negligible. a) Calculate the angular momentum of the projectile about the origin. b) Calculate the rate of change of this angular momentum. c) Calculate the torque acting on the projectile, about the origin, during its flight.

A couple is a set of two forces of equal magnitude and opposite directions, whose lines of action are parallel but not identical. Prove that the net torque of a couple of forces is independent of the pivot point about which the torque is calculated and of the points along their lines of action where the two forces are applied.

A space station is to provide artificial gravity to support long-term stay of astronauts and cosmonauts. It is designed as a large wheel, with all the compartments in the rim, which is to rotate at a speed that will provide an acceleration similar to that of terrestrial gravity for the astronauts (their feet will be on the inside of the outer wall of the space station and their heads will be pointing toward the hub). After the space station is assembled in orbit, its rotation will be started by the firing of a rocket motor fixed to the outer rim, which fires tangentially to the rim. The radius of the space station is \(R=50.0 \mathrm{~m},\) and the mass is \(M=2.40 \cdot 10^{5} \mathrm{~kg} .\) If the thrust of the rocket motor is \(F=1.40 \cdot 10^{2} \mathrm{~N},\) how long should the motor fire?

A force, \(\vec{F}=(2 \hat{x}+3 \hat{y}) \mathrm{N},\) is applied to an object at a point whose position vector with respect to the pivot point is \(\vec{r}=(4 \hat{x}+4 \hat{y}+4 \hat{z}) \mathrm{m} .\) Calculate the torque created by the force about that pivot point.
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