/*! 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 71 A solid cylindrical disk has a r... [FREE SOLUTION] | 91Ó°ÊÓ

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A solid cylindrical disk has a radius of 0.15 \(\mathrm{m}\). It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a \(45-N\) force is applied tangentially to the disk, perpendicular to the radius, the disk acquires an angular acceleration of \(120 \mathrm{rad} / \mathrm{s}^{2}\) What is the mass of the disk?

Short Answer

Expert verified
The mass of the disk is 5 kg.

Step by step solution

01

Identify Given Variables

We are given the following information: the radius of the disk is \( r = 0.15\, \mathrm{m} \), the applied force is \( F = 45\, \mathrm{N} \), and the angular acceleration is \( \alpha = 120\, \mathrm{rad/s}^2 \). We need to find the mass (\( m \)) of the disk.
02

Use Torque Formula

Torque (\( \tau \)) is given by the formula \( \tau = F \, \cdot \, r \), where \( F \) is the force applied tangentially, and \( r \) is the radius of the disk. Substitute the given values into the equation:\[ \tau = 45 \, \mathrm{N} \, \cdot \, 0.15 \, \mathrm{m} = 6.75 \, \mathrm{Nm} \]
03

Relate Torque to Angular Acceleration

The relationship between torque and angular acceleration is provided by the equation \( \tau = I \, \alpha \), where \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. The moment of inertia for a solid cylinder rotating about its center is \( I = \frac{1}{2} m r^2 \). Thus, the equation becomes:\[ 6.75 \, \mathrm{Nm} = \left(\frac{1}{2} m \cdot (0.15 \, \mathrm{m})^2\right) \cdot 120 \, \mathrm{rad/s}^2 \]
04

Solve for Mass

First, calculate the substitution inside the moment of inertia:\[ \frac{1}{2} \cdot (0.15)^2 = \frac{1}{2} \cdot 0.0225 = 0.01125 \]Substitute \[ 6.75 = m \cdot 0.01125 \cdot 120 \]Now solve for \( m \):\[ 6.75 = 1.35 m \]\[ m = \frac{6.75}{1.35} = 5 \]
05

Conclusion

The mass of the solid cylindrical disk is found to be \( 5 \, \mathrm{kg} \).

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Key Concepts

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

Torque
Torque is an essential concept in physics, particularly when dealing with rotational motion. It is the rotational equivalent of linear force and measures how much force is acting on an object to make it rotate. Torque is calculated using the formula:- \( \tau = F \times r \)- \( F \) is the force applied- \( r \) is the perpendicular distance from the pivot point (axis of rotation) to the line of action of the forceIn the context of the problem, when a force of 45 N is applied tangentially to the disk, torque helps us determine the twisting force responsible for the motion. It is important to ensure that the force is applied correctly (perpendicular to the radius) to maximize the produced torque.
Angular Acceleration
Angular acceleration is all about the rate of change of angular velocity over time. It tells us how quickly a rotating object speeds up or slows down its spin. Consider the equation:- \( \tau = I \times \alpha \)- \( \tau \) is torque- \( I \) is the moment of inertia- \( \alpha \) is angular accelerationThis equation shows that angular acceleration is directly proportional to torque and inversely proportional to the moment of inertia. In simple terms, the greater the torque applied to an object, the more quickly it accelerates unless it has a large moment of inertia. For our problem, the given angular acceleration of 120 rad/s² helps us calculate the moment of inertia and, subsequently, the mass of the disk.
Moment of Inertia
Moment of inertia quantifies how difficult it is to change the rotational motion of an object. It's similar to mass in linear motion, but for rotation. Every object has a unique moment of inertia based on its shape and distribution of mass. For a solid cylinder rotating about its central axis, such as our disk, the moment of inertia is:- \( I = \frac{1}{2} m r^2 \)- \( m \) is the mass of the object- \( r \) is the radiusUnderstanding the moment of inertia is crucial when you need to find how mass influences rotational movement. In the exercise, calculating this helps us ultimately determine the mass of the disk.
Solid Cylinder
A solid cylinder is a common shape in physics problems, especially those involving rotation. This shape, thanks to its symmetrical nature around its axis, makes it straightforward to calculate various physical quantities like torque, angular acceleration, and moment of inertia. Key characteristics of a solid cylinder include: - Uniform mass distribution - Revolves around its central axis When calculating the properties of a solid cylinder, remember the formulas associated with its geometry and consider the axis about which it rotates. These calculations allow us to determine quantities such as moment of inertia and leverage other properties to find the mass, as the problem required.

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Most popular questions from this chapter

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{c m}+M h^{2},\) where \(I_{c m}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A helicopter has two blades (see Figure 8.11 ); each blade has a mass of \(240 \mathrm{kg}\) and can be approximated as a thin rod of length \(6.7 \mathrm{m} .\) The blades are rotating at an angular speed of \(44 \mathrm{rad} / \mathrm{s}\). (a) What is the total moment of inertia of the two blades about the axis of rotation? (b) Determine the rotational kinetic energy of the spinning blades.

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of \(0.80 \mathrm{m}\) and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed \(v_{0,}\) so that the rod begins to rotate upward about the pivot. What must be the value of \(v_{0}\) such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{kg})\) is rotating at an angular velocity of 13.1 rad/s (see the drawing). The front brake is then applied for \(3.0 \mathrm{s},\) and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{m}\). The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9}\) J of energy. How fast would a \(13-k g\) flywhcel with a radius of \(0.30 \mathrm{m}\) have to rotate to store this much energy? Give your answer in rev/min.

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