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Chapter 11: Problem 40

Four Identical Particles Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

Short Answer

Expert verified
(a) 2.0 \text{ kg} \text{ m}^2, (b) 2.26 \text{ kg} \text{ m}^2, (c) 1.0 \text{ kg} \text{ m}^2

Step by step solution

01

Identify the Problem

The problem asks for the rotational inertia (moment of inertia) of a square-shaped arrangement of four particles about three different axes. Each particle has a mass of 0.50 kg, and each side of the square is 2.0 m.
02

Moment of Inertia Basics

The moment of inertia depends on the axis of rotation and is given by \[ I = \begin{cases} \frac{1}{12} M (a^2 + b^2) & \text{for a rod about its center} \ m r^2 & \text{for point masses} \end{cases} \] where \( m \) is the mass and \( r \) is the distance from the axis of rotation.
03

Axis Passing Through Midpoints of Opposite Sides (Part a)

For an axis passing through the midpoints of opposite sides and in the plane of the square, each particle is 1 m away from the axis. Using the formula for point masses: \[ I_a = 4 \times (0.50 \text{ kg} \times (1 \text{ m})^2) = 2.0 \text{ kg} \text{ m}^2 \]
04

Axis Through Midpoint of One Side and Perpendicular to Plane (Part b)

For an axis perpendicular to the plane of the square and passing through the midpoint of one of its sides, each particle is a different distance away. Two particles are 1 m away and two are \( \frac{\sqrt{5}}{2} \text{ m} \): \[ I_b = 2 \times (0.50 \text{ kg} \times 1 \text{ m}^2) + 2 \times (0.50 \text{ kg} \times (1.12 \text{ m})^2) = 2 \times 0.50 + 2 \times 0.63 = 2.26 \text{ kg} \text{ m}^2 \]
05

Axis Passing Through Two Diagonal Opposite Particles (Part c)

For an axis lying in the plane and passing through two diagonally opposite particles, the two particles on the axis contribute zero and the other two are \( \frac{2}{\sqrt{2}} = \frac{\sqrt{2}}{2} \text{ m} \) away from the axis: \[ I_c = 2 \times (0.50 \text{ kg} \times (\frac{2}{\text{ m}})^2) + 2 \times 0 = 0.50 \text{ kg} \text{ m}^2 \]

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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, also known as rotational inertia, is a measure of how difficult it is to change the rotational motion of an object. Think of it like rotational mass. It depends on both the mass of the object and how that mass is distributed relative to the chosen axis of rotation. For point masses, the formula to calculate the moment of inertia is \( I = \sum m_i r_i^2 \), where \( m_i \) is the mass of the particle and \( r_i \) is the distance from the axis of rotation. This distribution means that the farther the mass is from the axis, the larger the moment of inertia.
Axis of Rotation
The axis of rotation is an imaginary line that an object rotates around. The position of this axis greatly affects the moment of inertia. In the exercise, the rotational inertia is calculated for three different axes:
1. An axis through the midpoints of opposite sides and in the plane of the square.
2. An axis through the midpoint of one side and perpendicular to the plane of the square.
3. An axis passing through two diagonally opposite particles and lying in the plane of the square.
These different positions show how the distance of particles from the axis changes the moment of inertia.
Mass Distribution
Mass distribution refers to how mass is spread out in an object. In the given problem, mass is evenly distributed with four identical particles at the vertices of a square. When calculating the moment of inertia, this distribution is key as you have to consider how far each particle's mass is from the axis of rotation. For instance, in part (a), all particles are 1 m away from the axis, making the calculation straightforward. However, in parts (b) and (c), distances vary, making the calculations a bit more complex.
Rigid Body Dynamics
Rigid body dynamics deals with the motion of solid objects that do not deform during movement. This concept assumes the distance between particles does not change, simplifying our calculations. By treating the square arrangement of particles as a rigid body, we can straightforwardly calculate the moment of inertia for different axes. Understanding rigid body dynamics helps grasp how objects will rotate and respond to forces. It is foundational for studying systems in engineering, physics, and robotics.

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Most popular questions from this chapter

Vinyl Record A vinyl record on a turntable rotates at \(33 \frac{1}{3}\) rev/min. (a) What is its rotational speed in radians per second? What is the translational speed of a point on the record at the needle when the needle is (b) \(15 \mathrm{~cm}\) and (c) \(7.4 \mathrm{~cm}\) from the turntable axis?

Turntable Two A record turntable is rotating at \(33 \frac{1}{3}\) rev/min. A watermelon seed is on the turntable \(6.0 \mathrm{~cm}\) from the axis of rotation. (a) Calculate the translational acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction, \(\mu^{\text {stat }}\), between the seed and the turntable if the seed is not to slip? (c) Suppose that the turntable achieves its rotational speed by starting from rest and undergoing a constant rotational acceleration for \(0.25 \mathrm{~s}\). Calculate the minimum \(\mu^{\text {stat }}\) required for the seed not to slip during the acceleration period.

Flywheel Rotating A flywheel with a diameter of \(1.20 \mathrm{~m}\) has a rotational speed of 200 rev/min. (a) What is the rotational speed of the flywheel in radians per second? (b) What is the translational speed of a point on the rim of the flywheel? (c) What constant rotational acceleration (in revolutions per minute-squared) will increase the wheel's rotational speed to 1000 rev/min in 60 s? (d) How many revolutions does the wheel make during that \(60 \mathrm{~s}\) ?

A Disk A disk, initially rotating at \(120 \mathrm{rad} / \mathrm{s}\), is slowed down with a constant rotational acceleration of magnitude \(4.0 \mathrm{rad} / \mathrm{s}^{2} .(\mathrm{a})\) How much time does the disk take to stop? (b) Through what angle does the disk rotate during that time?

Meter Stick Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the \(20 \mathrm{~cm}\) mark. (Treat the stick as a thin rod.)
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