/*! 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 59 Two disks are rotating about the... [FREE SOLUTION] | 91Ó°ÊÓ

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Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s} .\) Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.4 \mathrm{rad} / \mathrm{s}\). The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

Short Answer

Expert verified
The moment of inertia of disk B is \(4.41 \, \text{kg} \cdot \text{m}^2\).

Step by step solution

01

Identify Known Values

We know the moment of inertia of disk A, \(I_A = 3.4 \, \text{kg} \cdot \text{m}^2\). The angular velocities before linking are \(\omega_A = +7.2 \, \text{rad/s}\) for disk A and \(\omega_B = -9.8 \, \text{rad/s}\) for disk B. After linking, the combined angular velocity is \(\omega_f = -2.4 \, \text{rad/s}\).
02

Apply Conservation of Angular Momentum

When no external torque acts on a system, the total angular momentum before and after an event remains constant. For this system, the angular momentum before linking must equal the angular momentum after linking. This can be expressed by the equation: \(I_A \omega_A + I_B \omega_B = (I_A + I_B) \omega_f\).
03

Substitute Known Values and Solve for \(I_B\)

Insert the known values into the conservation equation: \(3.4 \times 7.2 + I_B(-9.8) = (3.4 + I_B)(-2.4)\). Simplify and solve this equation for \(I_B\).
04

Simplify the Equation

Simplify \(3.4 \times 7.2 = 24.48\). Substitute this into the equation: \[24.48 - 9.8I_B = -8.16 - 2.4I_B\]. Then, gather all terms with \(I_B\) on one side and constant terms on the other.
05

Solve for \(I_B\)

Re-arrange the equation to \(24.48 + 8.16 = 9.8I_B - 2.4I_B\) which leads to \(32.64 = 7.4I_B\). Solve for \(I_B\) by dividing both sides by 7.4: \(I_B = \frac{32.64}{7.4} = 4.41 \, \text{kg} \cdot \text{m}^2\).

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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 is often referred to as the rotational equivalent of mass in linear motion. It measures how difficult it is to change the rotation of an object about a particular axis.
The moment of inertia depends on both the object's mass and its shape, as well as the axis about which it's rotating. - Formula: It's usually expressed in the form \( I = \sum m_i r_i^2 \), where \( m_i \) is the mass of each particle and \( r_i \) is the distance of each particle from the axis of rotation.- Units: The standard unit for moment of inertia in the International System of Units (SI) is \( ext{kg} \, \text{m}^2 \).In practice, calculating the moment of inertia can be straightforward for simple shapes like rods or spheres. However, it can be more complex for irregular objects. In exercises like the one above, you're often given the moment of inertia and must use it to solve problems related to rotational dynamics.
Angular Velocity
Angular velocity describes how fast an object is rotating. The concept is quite similar to linear velocity, but instead of moving through space, the object is rotating around an axis.
This can be either a point inside the object or a point outside the object in some rare cases. Angular velocity is a vector quantity, meaning it has both a magnitude and a direction. - Formula: If \( \theta \) is the angular displacement and \( t \) is the time taken, then the angular velocity \( \omega \) is \( \omega = \frac{d\theta}{dt} \).- Units: It is measured in radians per second (rad/s) in the SI system.Usually, problems like the one you've encountered will give you the angular velocities before and after a certain event, allowing you to apply the conservation of momentum principles for rotational motion. It helps us determine the resulting angular velocity or other quantities like the moment of inertia.
Rotational Dynamics
Rotational dynamics covers the forces and torques and their effects on rotation. Just as forces cause changes in linear motion, torques produce changes in rotational motion. Understanding rotational dynamics requires grasping how torques interact with objects' moments of inertia.
- Formula: The core equation linking these concepts is Newton's second law for rotation, \( \tau = I \alpha \), where \( \tau \) is torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.- Conservation: A critical concept in rotational dynamics is the conservation of angular momentum, which states that if no external torques act on a system, its angular momentum remains constant. This principle is perfectly demonstrated in your exercise, where the total angular momentum before and after the disks are linked remains the same.Rotational dynamics allow us to predict how rotating bodies will behave under various conditions, which is essential for understanding both theoretical physics and practical engineering problems.

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Most popular questions from this chapter

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a $$30.0^{\circ}$$ angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

A 9.75-m ladder with a mass of 23.2 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of \(245 \mathrm{N}\). At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of \(1.80 \mathrm{rad} / \mathrm{s}^{2}\) about an axis passing through the bottom end of the ladder. The ladder's center of gravity lics halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m},\) an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{kg} \cdot \mathrm{m}^{2} .\) A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{s} .\) (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

Two spheres are each rotating at an angular speed of \(24 \mathrm{rad} / \mathrm{s}\) about axes that pass through their centers. Each has a radius of \(0.20 \mathrm{m}\) and a mass of \(1.5 \mathrm{kg}\). However, as the figure shows, one is solid and the other is a thin-walled spherical shell. Suddenly, a net external torque due to friction (magnitude \(=0.12 \mathrm{N} \cdot \mathrm{m}\) ) begins to act on each sphere and slows the motion down. Concepts: (i) Which sphere has the greater moment of inertia and why? (ii) Which sphere has the angular acceleration (a deceleration) with the smaller magnitude? (iii) Which sphere takes a longer time to come to a halt? Calculations: How long does it take each sphere to come to a halt?

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.

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