/*! 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 69 The moment of inertia of the emp... [FREE SOLUTION] | 91Ó°ÊÓ

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The moment of inertia of the empty tumtable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m}\). the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s}\). What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. A. \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) B. \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) C. \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) D. \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

Short Answer

Expert verified
The person's moment of inertia is \(6.0 \, \mathrm{kg} \cdot \mathrm{m}^2\).

Step by step solution

01

Determine Angular Acceleration

First, we need to calculate the angular acceleration \( \alpha \) using the formula relating torque \( \tau \), moment of inertia \( I \), and angular acceleration: \( \tau = I_{total} \cdot \alpha \).Given \( \tau = 2.5 \, \mathrm{N} \cdot \mathrm{m} \) and the angular speed \( \omega = 1.0 \, \mathrm{rad/s} \) achieved in \( 3.0 \, \mathrm{s} \), we can find \( \alpha \) using the relation: \( \omega = \alpha \cdot t \).Thus, \( \alpha = \frac{\omega}{t} = \frac{1.0 \, \mathrm{rad/s}}{3.0 \, \mathrm{s}} = \frac{1}{3} \, \mathrm{rad/s^2} \).
02

Calculate Total Moment of Inertia

Using \( \tau = I_{total} \cdot \alpha \), substitute \( \tau = 2.5 \, \mathrm{N} \cdot \mathrm{m} \) and \( \alpha = \frac{1}{3} \, \mathrm{rad/s^2} \):\[ I_{total} = \frac{\tau}{\alpha} = \frac{2.5}{\frac{1}{3}} = 7.5 \, \mathrm{kg} \cdot \mathrm{m}^2 \].
03

Subtract Moment of Inertia of Turntable

The total moment of inertia \( I_{total} = 7.5 \, \mathrm{kg} \cdot \mathrm{m}^2 \) includes both the turntable and the person. Given the turntable's moment of inertia is \( 1.5 \, \mathrm{kg} \cdot \mathrm{m}^2 \), subtract it from \( I_{total} \) to find the person's contribution.So, \( I_{person} = 7.5 - 1.5 = 6.0 \, \mathrm{kg} \cdot \mathrm{m}^2 \).

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Key Concepts

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

Angular Acceleration
Angular acceleration is essentially how quickly something speeds up or slows down during rotational motion. It's similar to linear acceleration, but instead of dealing with speed in straight lines, we're talking about changes in angles over time. When an object rotates faster or slower, it's experiencing angular acceleration. In this exercise, we're given the final angular speed and the time taken to reach from rest to that speed. We calculate the angular acceleration \(\alpha\) using the formula \(\omega = \alpha \cdot t\). Here, \(\omega\) is the angular speed and \(t\) is the time period. By plugging in the numbers from the problem, \(\omega = 1.0 \, \text{rad/s}\) and \(t = 3.0 \, \text{s}\), we find \(\alpha = \frac{1}{3} \, \text{rad/s}^2\). This shows how the object's rotational speed changes over each second. Understanding angular acceleration is key to analyzing all kinds of spinning or rotating objects, from wheels to planets.
Torque
Torque is what causes objects to start rotating. It's the rotational counterpart to force. Whenever you twist or turn something, you're applying torque. In this exercise, we have a constant torque of \(2.5 \, \text{N} \cdot \text{m}\). This torque is responsible for increasing the angular speed of the turntable-person system.
Just like force causes linear acceleration when you push or pull an object, torque causes angular acceleration. The relation between torque, moment of inertia, and angular acceleration is given by the formula \(\tau = I_{total} \cdot \alpha\). Here, \(\tau\) represents the torque, and \(I_{total}\) is the total moment of inertia of the system, combining the turntable and the person. Understanding torque helps us comprehend how various forces lead to rotational movements.
Rotational Motion
Rotational motion is everywhere in our daily lives. Anytime an object spins or turns around a point, it is experiencing rotational motion. This type of motion is characterized by variables like angular speed, angular acceleration, and moment of inertia.
In the problem, the turntable system transitions from rest to a spinning state due to applied torque, illustrating how rotational motion is initiated and analyzed. The moment of inertia plays a crucial role here, as it determines how much torque is needed to change the angular speed. In our example, the system's total moment of inertia is calculated, showing how it influences the overall motion. When we talk about rotational motion, we break it down into components like angular distance traveled, the torque causing it, and the resistance faced by the rotating body due to its moment of inertia.
Physics Problem Solving
Solving physics problems often involves breaking down a complex situation into more manageable parts. For rotational motion problems like this one, it's about identifying the right equations and relationships.
  • First, identify the given quantities, such as torque, time, and angular speed.
  • Use relevant formulas like \(\omega = \alpha \cdot t\) and \(\tau = I_{total} \cdot \alpha\).
  • Calculate quantities step by step, like finding angular acceleration first.
  • Piece these calculations together to find the desired unknowns, such as an individual moment of inertia.
This organized approach makes it easier to solve complex physics problems. By understanding concepts like angular acceleration and torque, and knowing which formulas to apply, we can systematically find solutions to various problems, making physics not just more manageable but also more enjoyable.

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Most popular questions from this chapter

II Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a newtron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform. solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{~km}\) (comparable to our sun); its final radius is \(16 \mathrm{~km}\). If the original star rotated once in 30 days, find the angular speed of the neutron star.

A grindstone in the shape of a solid disk with diameter \(0.520 \mathrm{~m}\) and a mass of \(50.0 \mathrm{~kg}\) is rotating at \(850 \mathrm{rev} / \mathrm{min}\). You press an axe against the rim with a normal force of \(160 \mathrm{~N}\) (see Figure 10.50 ). and the grindstone comes to rest in \(7.50 \mathrm{~s}\). Find the coefficient of kinctic friction befween the axe and the grindstone. There is negligible friction in the bearings.

A solid uniform sphere of mass \(5 \mathrm{~kg}\) and radius \(0.1 \mathrm{~m}\) rolls down an inclined plane without slipping. If the sphere's center of mass moves through a vertical distance of \(1.5 \mathrm{~m}\), what is the rotational work done on the sphere by the static-friction force?

If the competitor leans slightly farther back (increasing the angle between his body and the vertical) but remains stationary in this new position, which statement is true? Assume that the rope remains horizontal. A. The difference between \(T_{1}\) and \(T_{2}\) increases, balancing the increased torque about his feet that his weight produces when he leans farther back. B. The difference between \(T_{1}\) and \(T_{2}\) decreases, balancing the increased torque about his feet that his weight produces when he leans farther back. C. Neither \(T_{1}\) nor \(T_{2}\) changes because no other forces are changing. D. Both \(T_{1}\) and \(T_{2}\) change, but the difference between them remains the same.

Two people are carrying a uniform wooden board that is \(3.00 \mathrm{~m}\) long and weighs \(160 \mathrm{~N}\). If one person applies an upward force equal to \(60 \mathrm{~N}\) at one end, at what point and with what force does the other person lift? Start with a free-body diagram of the board.

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