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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 kg m², (b) 1.0 kg m², (c) 2.0 kg m².

Step by step solution

01

Understanding Rotational Inertia

Rotational inertia, also known as the moment of inertia, depends on how mass is distributed relative to the axis of rotation. It is given by the formula \( I = \sum m r^2 \), where \( m \) is the mass of a particle and \( r \) is the distance from the axis of rotation to the particle.
02

Calculate for Axis A

For part (a), the axis passes through the midpoints of opposite sides in the plane of the square. Each particle is \(1\mathrm{~m}\) from this axis. Therefore, the rotational inertia \( I \) is:\[ I = 4 \times (0.5 \mathrm{~kg}) \times (1 \mathrm{~m})^2 = 2.0 \mathrm{~kg} \, \mathrm{m}^2 \]
03

Calculate for Axis B

For part (b), the axis passes through the midpoint of one of the sides and is perpendicular to the plane. Two particles are on the axis, making \( r = 0 \) for them. The other two are \(1\mathrm{~m}\) away. The rotational inertia \( I \) is:\[ I = 2 \times (0.5 \mathrm{~kg}) \times (1 \mathrm{~m})^2 = 1.0 \mathrm{~kg} \, \mathrm{m}^2 \]
04

Calculate for Axis C

For part (c), the axis passes through two diagonally opposite particles. Each of these particles is \(0\mathrm{~m}\) from the axis, but the other two are along the diagonal, at \( \sqrt{2} \mathrm{~m} \) from the axis. The rotational inertia \( I \) is:\[ I = 2 \times (0.5 \mathrm{~kg}) \times (\sqrt{2} \mathrm{~m})^2 = 2 \times (0.5 \mathrm{~kg}) \times 2 \mathrm{~m}^2 = 2.0 \mathrm{~kg} \, \mathrm{m}^2 \]

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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 fundamental concept in rotational dynamics. It is akin to mass in linear motion but applies to rotational motion. Often denoted as \( I \), it quantifies how a body's mass is distributed relative to a chosen axis of rotation. The moment of inertia takes into account both the mass and the square of the distance each mass element is from the axis. This relationship is captured by the formula:
  • \( I = \sum m r^2 \)
Here, \( m \) is the mass of a particle and \( r \) is its distance to the axis of rotation.
The larger the moment of inertia, the harder it is to change the rotational speed of the object. This is because more significant rotational inertia means more mass is far from the axis, requiring more torque to achieve the same angular acceleration.
Axis of Rotation
The axis of rotation plays a crucial role in determining an object's moment of inertia. It is an imaginary line around which an object rotates and can adopt various orientations based on the scenario. For the square in the discussed exercise, different axes of rotation yield different moments of inertia:
  • In-plane axis through midpoints of opposite sides: This axis equally divides the square. Particles' distance from the axis governs their contribution to rotational inertia.
  • Perpendicular axis through a side's midpoint: Here, half the particles lie directly on the axis, resulting in \( r = 0 \) for those two.
  • Axis along diagonally opposite particles: This axis crosses two particles directly, simplifying their individual contributions to zero rotational distance.
Understanding the axis is crucial because the calculated moment of inertia changes with different axes due to varying distances \( r \) in the inertia formula.
Mass Distribution
Mass distribution significantly affects the rotational inertia of a body. It refers to how the mass is spread out in relation to the axis of rotation.
In our example, the square's particles—each with mass \(0.5 \mathrm{~kg}\)—are placed at its vertices, affecting their distance from different axes. Here’s how mass positioning impacts inertia:
  • The further a mass is from the axis, greater is its contribution to the moment of inertia, posing higher resistance to rotational changes.
  • Uniform mass distribution, like in the square, allows application of symmetry principles, simplifying calculations for different rotational axes.
Thus, analyzing mass distribution in relation to the chosen axis is fundamental when determining an object's moment of inertia.
Rigid Body Dynamics
Rigid body dynamics deals with the motion of solid objects that do not deform during movement.
In this field, an object's entire shape and structure move without internal displacement of particles. An important consideration is that a rigid body's behavior under rotation can be predicted using rotational inertia and external forces. For example, in the given problem:
  • The square, though composed of discrete masses (particles tied by rods), is treated as one rigid entity.
  • Such simplification is valid because the rods do not bend, and the distances between the masses do not change.
By analyzing the square's rotational inertia about different axes, we can understand how the entire structure behaves when subjected to rotational motion. Rigid body dynamics allows us to model and predict this behavior efficiently.

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Most popular questions from this chapter

A thin rod of length \(0.75 \mathrm{~m}\) and mass \(0.42 \mathrm{~kg}\) is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed \(4.0 \mathrm{rad} / \mathrm{s} .\) Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just before it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.)

At \(t=0,\) a flywheel has an angular velocity of \(4.7 \mathrm{rad} / \mathrm{s}, \mathrm{a}\) constant angular acceleration of \(-0.25 \mathrm{rad} / \mathrm{s}^{2},\) and a reference line at \(\theta_{0}=0 .\) (a) Through what maximum angle \(\theta_{\max }\) will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at \(\theta=\frac{1}{2} \theta_{\max } ?\) At what (d) negative time and (e) positive time will the reference line be at \(\theta=10.5\) rad? (f) Graph \(\theta\) versus \(t,\) and indicate your answers.

A car starts from rest and moves around a circular track of radius \(30.0 \mathrm{~m} .\) Its speed increases at the constant rate of \(0.500 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the magnitude of its net linear acceleration 15.0 s later? (b) What angle does this net acceleration vector make with the car's velocity at this time?

A golf ball is launched at an angle of \(20^{\circ}\) to the horizontal, with a speed of \(60 \mathrm{~m} / \mathrm{s}\) and a rotation rate of \(90 \mathrm{rad} / \mathrm{s} .\) Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

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