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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 magnitude of the angular momentum of the asssembly?

Short Answer

Expert verified
(a) 0.0009936 kg m虏, (b) 0.00028152 kg m虏/s, (c) 0.00084456 kg m虏/s.

Step by step solution

01

Rotational Inertia of the Assembly

To find the rotational inertia (moment of inertia) of the assembly around point O, we consider the rotation of each particle about point O. The moment of inertia for a single particle is given by \( I = m r^2 \), where \( m \) is the mass of the particle and \( r \) is the perpendicular distance from the rotation axis to the particle.1. Each particle has a mass \( m = 23 \text{ g} = 0.023 \text{ kg} \) and the length of each rod is \( d = 12 \text{ cm} = 0.12 \text{ m} \).2. The rigid setup suggests that the particles form an equilateral triangle with point O.3. Hence, all particles are at the same distance \( d = 0.12 \text{ m} \) from point O.The rotational inertia for the entire assembly is:\[I_{total} = 3 imes m imes d^2 = 3 imes 0.023 imes (0.12)^2 = 3 imes 0.023 imes 0.0144 = 0.0009936 ext{ kg} ext{ m}^2.\]
02

Angular Momentum of the Middle Particle

The angular momentum \( L \) of a particle is calculated using the formula \( L = I \omega \), where \( I \) is the moment of inertia of the particle and \( \omega \) is the angular speed.For a single particle at distance \( r = d = 0.12 \text{ m} \) from the rotation axis:\[I_{particle} = m imes r^2 = 0.023 imes (0.12)^2 = 0.023 imes 0.0144 = 0.0003312 ext{ kg} ext{ m}^2.\]The angular momentum of the middle particle is:\[L_{particle} = I_{particle} \times \omega = 0.0003312 \times 0.85 = 0.00028152 ext{ kg} ext{ m}^2/ ext{s}.\]
03

Angular Momentum of the Assembly

For three particles forming the assembly, consider their combined angular momentum using the individual moments of inertia:1. The total angular momentum is the sum of the angular momentum of each particle.Total angular momentum for the assembly:\[L_{total} = I_{total} \times \omega = 0.0009936 \times 0.85 = 0.00084456 ext{ kg} ext{ m}^2/ ext{s}.\]

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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 crucial concept in rotational dynamics, providing an understanding of rotational motion. It is analogous to linear momentum, but applies to rotating systems. Angular momentum (\(L\)) is determined using the formula:
  • \(L = I \cdot \omega\)
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. In this exercise, each particle contributes to the total angular momentum of the system.

The angular momentum helps in analyzing how rotational effects, like spinning, will change over time. The direction of angular momentum depends on the direction of the axis of rotation. Therefore, it not only tells us how fast something spins but also in which direction.

Angular momentum is conserved in isolated systems, meaning the total angular momentum remains constant if no external torques act on the system. This property is vital in mechanics and can often simplify complex problems.
Moment of Inertia
The moment of inertia, often denoted as \(I\), is a property of a physical object that quantifies its resistance to rotational motion around an axis. It depends on both the mass of the object and how that mass is distributed with respect to the axis of rotation.

For a point mass, the formula for moment of inertia is:
  • \(I = m \cdot r^2\)
where \(m\) is the mass of the particle and \(r\) is the distance from the axis of rotation.

In this problem, the total moment of inertia of the assembly consists of three particles at equal distances, creating an equilateral triangle. The rotational inertia of the system was calculated by considering each particle鈥檚 contribution. Because the rods are considered to have negligible mass, only the mass of the particles affects the moment of inertia.

Moment of inertia plays a crucial role in dynamics as it influences how much torque is needed for a certain angular acceleration. The distribution of mass significantly affects rotational stability and dynamics.
Equilateral Triangle
An equilateral triangle is a special type of triangle where all three sides are equal in length. This uniformity also means that all interior angles are equal, each being 60 degrees. In the context of rotation problems, an equilateral triangle can offer symmetry, simplifying calculations.

In this exercise, the particles and point \(O\) form an equilateral triangle. This configuration ensures each particle is equidistant from \(O\), making the computations of both moment of inertia and angular momentum more straightforward.

Proper understanding of a geometric shape like an equilateral triangle helps when setting up problems in physics, especially those dealing with symmetry and rotational properties. Such symmetries often allow the problem to reduce to simpler calculations, saving time and effort while applying principles like uniform distribution of mass or other symmetrical properties.

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Most popular questions from this chapter

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg}\). Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go- round and child after the child has jumped onto the merry-go-round.

At time \(t=0\), a \(2.0 \mathrm{~kg}\) particle has the position vector \(\vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) relative to the origin. Its velocity is given by \(\vec{v}=\left(-6.0 t^{2} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}\) for \(t \geq 0\) in seconds. About the origin, what are (a) the particle's angular momentum \(\vec{L}\) and \((\mathrm{b})\) the torque \(\vec{\tau}\) acting on the particle, both in unit-vector notation and for \(t>0 ?\) About the point \((-2.0 \mathrm{~m},-3.0 \mathrm{~m}, 0)\), what are (c) \(\vec{L}\) and (d) \(\vec{\tau}\) for \(t>0 ?\)

A solid brass ball of mass \(0.280 \mathrm{~g}\) will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius \(R=14.0 \mathrm{~cm}\), and the ball has radius \(r \ll R\). (a) What is \(h\) if the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height \(h=\) \(6.00 R\), what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball at point \(Q\) ?

A plum is located at coordinates \((-2.0 \mathrm{~m}, 0,4.0 \mathrm{~m}) .\) In unit- vector notation, what is the torque about the origin on the plum if that torque is due to a force \(\vec{F}\) whose only component is (a) \(F_{x}=\) \(6.0 \mathrm{~N},(\mathrm{~b}) F_{x}=-6.0 \mathrm{~N},(\mathrm{c}) F_{z}=6.0 \mathrm{~N}\), and \((\mathrm{d}) F_{z}=-6.0 \mathrm{~N} ?\)

In unit-vector notation, what is the torque about the origin on a jar of jalape帽o peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}\), \(4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}+(5.0 \mathrm{~N}) \hat{\mathrm{k}},(\mathrm{b})\) force \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}-(5.0 \mathrm{~N}) \hat{\mathrm{k}}\), and \((\mathrm{c})\) the vector sum of \(\vec{F}_{1}\) and \(\vec{F}_{2} ?(\mathrm{~d})\) Repeat part \((\mathrm{c})\) for the torque about the point with coordinates \((3.0 \mathrm{~m}, 2.0 \mathrm{~m}, 4.0 \mathrm{~m})\).
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