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Chapter 11: Problem 37

Three Particles of mass \(m=23 \mathrm{~g}\) are fastened to three rods of length \(d=12 \mathrm{~cm}\) and negligible mass. The rigid assembly rotates around point \(O\) at the angular speed \(\omega=0.85 \mathrm{rad} / \mathrm{s} .\) About \(O\) what are (a) the rotational inertia of the assembly, (b) the magnitude of the angular momentum of the middle particle, and (c) the mag. nitude of the angular momentum of the asssembly?

Short Answer

Expert verified
(a) 0.003456 kg路m虏, (b) 0.00028152 kg路m虏/s, (c) 0.0029376 kg路m虏/s.

Step by step solution

01

Understanding Rotational Inertia

The rotational inertia (moment of inertia) of multiple particles concerning a point is given by the sum of each particle's mass times the square of its distance from the rotation axis. Given three particles at the ends of rods of length \(d = 12\, \text{cm} = 0.12 \text{ m}\), and each with mass \(m = 23\, \text{g} = 0.023\, \text{kg}\), the rotational inertia \(I\) can be calculated for each particle about point \(O\). Since all particles are equidistant from point \(O\), the equation for rotational inertia is \(I = 3m d^2\).
02

Calculate Rotational Inertia

Substitute the values into the equation. \[I = 3 \times 0.023 \times (0.12)^2 = 3 \times 0.023 \times 0.0144 = 0.09936 \times 10^{-3} \text{ kg} \cdot \text{m}^2 \]. Therefore, \(I = 0.003456 \text{ kg} \cdot \text{m}^2\).
03

Understanding Angular Momentum of Middle Particle

The angular momentum \(L\) of a single particle is given by \(L = I_p \omega\) where \(I_p = md^2\) is the moment of inertia of that particle about point \(O\) and \(\omega = 0.85\, \mathrm{rad/s}\).
04

Calculate Angular Momentum of Middle Particle

For the middle particle: \[L = 0.023 \times (0.12)^2 \times 0.85 \]. Calculate the moment of inertia of that particle \(I_p = 0.023 \times 0.0144 = 0.03312 \times 10^{-3}\). Then, \(L = 0.0003312 \times 0.85 = 0.00028152 \text{ kg} \cdot \text{m}^2/\text{s}\).
05

Understanding Total Angular Momentum of Assembly

The total angular momentum of the assembly around point \(O\) can be calculated using \(L = I \omega\) for the entire system's rotational inertia as calculated previously.
06

Calculate Total Angular Momentum of the Assembly

Substitute the calculated rotational inertia and angular speed into the formula: \[L = 0.003456 \times 0.85 = 0.0029376\, \text{kg} \cdot \text{m}^2/\text{s}\].

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Key Concepts

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

Angular Momentum
Angular momentum is a measure of the rotational motion of an object. It combines the object's moment of inertia and its angular velocity to give a sense of how much motion is conserved when a force is applied. For a single particle, the angular momentum \(L\) is given by the formula \(L = I_p \cdot \omega\), where \(I_p\) is the individual particle's moment of inertia and \(\omega\) is the angular velocity. In the context of rotating systems, angular momentum provides a powerful way to predict how objects will behave under changes in motion and forces.
  • Conservation of Angular Momentum: Angular momentum is conserved in a closed system, meaning the total angular momentum remains constant unless acted upon by an external torque.
  • Units: The units of angular momentum are kilogram meter squared per second (kg路m虏/s).
  • Applications: Angular momentum is crucial in understanding phenomena ranging from spinning ice skaters to the rotation of planets.
Moment of Inertia
The moment of inertia, often referred to as rotational inertia, measures an object's resistance to changes in its rotational motion. It is essentially the rotational equivalent of mass in linear motion.
For point masses, the moment of inertia \(I\) about a given axis can be calculated using the formula \(I = \, \Sigma m_i r_i^2\), where \(m_i\) is the mass of each particle, and \(r_i\) is the distance of each mass from the axis of rotation. This formula implies that the further away the mass is from the axis of rotation, the higher the moment of inertia, making it harder to change the rotational motion.
  • Distribution of Mass: Moment of inertia depends heavily on how mass is distributed relative to the axis of rotation.
  • Practical Calculation: In mathematical problems, when dealing with multiple identical point masses equidistant from a pivot, the equation simplifies to \(I = nm \, d^2\), where \(n\) is the number of masses.
  • Units: Measured in kilogram meter squared (kg路m虏).
Rigid Body Dynamics
Rigid body dynamics involves the study of motion of solid objects that do not deform during movement. The rigid body assumption simplifies the analysis of motions involving rotation, translation, or both, making it a crucial concept in understanding mechanical systems.
Rigid bodies have fixed shapes and sizes, and calculations regarding rotation often use the concepts of angular momentum and moment of inertia. This helps in deducing how the body will react under various forces and torques.
  • Rotation and Translation: A rigid body can either translate (move from one place to another without rotating), rotate (spin about an axis), or do both simultaneously.
  • Equilibrium: For a rigid body in equilibrium, both the sum of forces and the sum of torques must be zero.
  • Applications: Rigid body dynamics forms the foundation for analyzing systems like machinery, vehicles, and even human motion in biomechanics.

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Most popular questions from this chapter

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.2 \mathrm{~mm},\) and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) linear speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) angular speed?

Shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m},\) its inner radius \(R_{1}\) is \(R_{2} / 2.00\) its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1} ?\)

A constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(10 \mathrm{~N}\) is applied to a wheel of mass \(10 \mathrm{~kg}\) and radius \(0.30 \mathrm{~m}\). The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude \(0.60 \mathrm{~m} / \mathrm{s}^{2}\). (a) In unit-vector notation, what is the frictional force on the wheel? (b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

A uniform block of granite in the shape of a book has face dimensions of \(20 \mathrm{~cm}\) and \(15 \mathrm{~cm}\) and a thickness of \(1.2 \mathrm{~cm} .\) The density (mass per unit volume) of granite is \(2.64 \mathrm{~g} / \mathrm{cm}^{3} .\) The block rotates around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that axis is \(0.104 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its rotational kinetic energy about that axis?

A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0,\) a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?
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