/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A thin uniform bar has two small... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin uniform bar has two small balls glued to its ends. The bar is 2.00 \(\mathrm{m}\) long and has mass \(4.00 \mathrm{kg},\) while the balls each have mass 0.500 \(\mathrm{kg}\) and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes: (a) an axis perpendicular to the bar through its center; (b) an axis perpendicular to the bar through one of the balls; (c) an axis parallel to the bar through both balls.

Short Answer

Expert verified
(a) 2.33 kg m²; (b) 7.33 kg m²; (c) 0.00 kg m².

Step by step solution

01

Moment of Inertia about center (part a)

To find the moment of inertia about an axis through the center of the bar, use the formula:\[ I = I_{bar} + I_{balls} \]The moment of inertia of the bar about its center is given by:\[ I_{bar} = \frac{1}{12} m L^2 \]where \( m = 4.00 \, \text{kg} \) and \( L = 2.00 \, \text{m} \). Thus, \[ I_{bar} = \frac{1}{12} (4.00) (2.00)^2 = \frac{1}{12} (16.00) = 1.33 \, \text{kg} \, \text{m}^2 \]For the balls, treat them as point masses at a distance \( L/2 = 1.00 \, \text{m} \) from the axis, so:\[ I_{balls} = 2 \times m_{ball} \times \left( \frac{L}{2} \right)^2 = 2 \times 0.500 \times (1.00)^2 = 1.00 \, \text{kg} \, \text{m}^2 \]So, the total moment of inertia:\[ I = 1.33 + 1.00 = 2.33 \, \text{kg} \, \text{m}^2 \]
02

Moment of Inertia about one of the balls (part b)

For an axis through one of the balls, the moment of inertia will only be the sum of the bar (treated as if one end is at the axis) and the mass of the other ball at a full length distance \(L = 2.00 \, \text{m}\):\[ I = I_{bar, \, ext{end}} + I_{ball} \]The bar's moment of inertia about its end:\[ I_{bar, \, ext{end}} = \frac{1}{3} m L^2 = \frac{1}{3} \times 4.00 \times (2.00)^2 = \frac{1}{3} \times 16.00 = 5.33 \, \text{kg} \, \text{m}^2 \]For the ball on the opposite end, full distance \(L = 2.00 \, \text{m}\):\[ I_{ball} = 0.500 \times (2.00)^2 = 2.00 \, \text{kg} \, \text{m}^2 \]So, the moment of inertia:\[ I = 5.33 + 2.00 = 7.33 \, \text{kg} \, \text{m}^2 \]
03

Moment of Inertia parallel to the bar (part c)

For an axis parallel to the bar, the moment of inertia is simply the sum of the two point masses since the bar contributes nothing because it's on the axis.The moment of inertia for each ball:\[ I_{ball} = m_{ball} \times (0.00)^2 = 0.00 \, \text{kg} \, \text{m}^2 \]Therefore, the total moment of inertia is:\[ I = 2 \times 0.00 = 0.00 \, \text{kg} \, \text{m}^2 \]This accounts for the displacement parallel, which contributes no inertia.

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Key Concepts

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

Rotational Dynamics
Rotational dynamics studies the motion of objects that rotate around an axis, encompassing concepts such as torque, angular momentum, and moment of inertia. It is analogous to linear dynamics, but instead of dealing with straight-line motion, it focuses on rotational motion.
Understanding rotational dynamics is crucial when analyzing systems where components spin or rotate, such as wheels, gears, or celestial bodies. The key equation in rotational dynamics is Newton's second law for rotation:
  • \( \tau = I \alpha \)
Here, \(\tau\) is the net torque applied to the body, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. This relationship highlights how the distribution of mass (moment of inertia) affects an object's rotational acceleration.
  • Torque is the rotational equivalent of force.
  • Angular momentum is the rotational equivalent of linear momentum.
These concepts are foundational for understanding complex systems such as engines and turbines.
Axis of Rotation
The axis of rotation is an imaginary line around which an object rotates. The characteristics of this axis, such as its position and if it is fixed or moving, greatly influence the mechanics of the system. An object's moment of inertia, and thus its rotational dynamics, depends significantly on this axis.
In the exercise, the moment of inertia is calculated about different axes:
  • Through the center of the bar, perpendicular to its length.
  • Through one of the balls, also perpendicular.
  • Parallel to the bar and through the balls.
Each axis orientation changes the distribution of mass relative to the axis, altering the moment of inertia. An axis through the center may require the entire mass to be considered equally distributed, while an axis at one end will result in different calculations due to asymmetrical distribution.
Point Masses
Point masses simplify calculations by treating objects as single mass points. This is useful when an object's size is negligible compared to the distances involved. In the provided exercise, the balls at the ends of the bar represent point masses. This assumption allows us to easily apply the formula for moment of inertia, which for a point mass is:
  • \( I = m r^2 \)
Here, \(m\) is the mass of the point and \(r\) is the distance from the axis of rotation. The moment of inertia for a point mass increases as the square of the distance from the axis, emphasizing how shifting mass away from the axis increases resistance to rotation. This is why positioning of mass can be critical in machinery and structure design.
Parallel Axis Theorem
The parallel axis theorem is a useful tool in rotational dynamics when you need to find the moment of inertia of a body about an axis that is parallel to, but not coincident with, an axis through the center of mass. The theorem states:
  • \( I = I_{cm} + Md^2 \)
In this formula, \(I_{cm}\) is the moment of inertia about the center of mass axis, \(M\) is the total mass of the object, and \(d\) is the distance between the two parallel axes.
The theorem makes it easier to calculate the moments of inertia around various points, which is particularly useful in composite systems. In the exercise example, calculating moments of inertia through various axes requires shifting from the center of mass to other axes—this is precisely where the parallel axis theorem comes into play. It helps poise how changes in axis location affect rotational behavior.

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Most popular questions from this chapter

\(\bullet\) Electric drill. According to the shop manual, when drilling a 12.7 -mm-diameter hole in wood, plastic, or aluminum, a drill should have a speed of 1250 rev/min. For a 12.7 - -mm-diameter drill bit turning at a constant 1250 \(\mathrm{rev} / \mathrm{min}\) , find (a) the maximum linear speed of any part of the bit and (b) the maximum radial acceleration of any part of the bit.

A solid uniform spherical stone starts moving from rest at the top of a hill. At the bottom of the hill the ground curves upward, launching the stone vertically a distance \(H\) below its start. How high will the stone go (a) if there is no friction on the hill and (b) if there is enough friction on the hill for the stone to roll without slipping? (c) Why do you get two different answers even though the stone starts with the same gravitational potential energy in both cases?

\(\bullet\) (a) Calculate the angular velocity (in rad/s) of the second, minute, and hour hands on a wall clock. (b) What is the period of each of these hands?

\(\bullet\) (a) What angle in radians is subtended by an arc 1.50 \(\mathrm{m}\) in length on the circumference of a circle of radius 2.50 \(\mathrm{m} ?\) What is this angle in degrees? (b) An arc 14.0 \(\mathrm{cm}\) in length on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 \(\mathrm{m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

\(\bullet\) The kinetic energy of walking. If a person of mass \(M\) simply moved forward with speed \(V,\) his kinetic energy would be \(\frac{1}{2} M V^{2} .\) However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13\(\%\) of a person's mass, while the legs and feet together account for 37\(\% .\) For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm 30^{\circ}\) (a total of \(60^{\circ}\) ) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than beingbent, which is not quite true. Let us consider a 75 kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) long. (a) What is the avertage angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

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