/*! 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 46 The parallel axis theorem provid... [FREE SOLUTION] | 91Ó°ÊÓ

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The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+M h^{2},\) where \(I_{\mathrm{cm}}\) 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.

Short Answer

Expert verified
The moment of inertia is \( I = \frac{3}{2} M R^2 \).

Step by step solution

01

Identify known components

We begin with what is given in the problem. The cylinder in question has a radius, \( R \), and the axis of rotation is on the surface of the cylinder, perpendicular to its ends. To apply the parallel axis theorem, we need \( I_{\text{cm}} \), the moment of inertia through the center of mass, \( M \), the mass of the cylinder, and \( h \), the distance between the axis through the center and the axis of interest (which is the radius in this case).
02

Determine known moment of inertia through center of mass

For a solid cylinder with its axis through the center along its length, the moment of inertia is given by \( I_{\text{cm}} = \frac{1}{2} M R^2 \), where \( M \) is the mass of the cylinder and \( R \) is its radius.
03

Apply parallel axis theorem formula

The parallel axis theorem states that the moment of inertia about any axis is \( I = I_{\text{cm}} + M h^{2} \). For this case, \( I_{\text{cm}} = \frac{1}{2} M R^2 \) and the distance \( h = R \) (since the axis is at the surface, one radius away from the center). Substitute these values into the equation to get \( I = \frac{1}{2} M R^2 + M R^2 \).
04

Simplify the expression

Simplify the expression: \[ I = \frac{1}{2} M R^2 + M R^2 = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \].
05

Final Expression

The expression for the moment of inertia of the cylinder with respect to the given axis is \( I = \frac{3}{2} M R^2 \).

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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 physics. It measures how much resistance an object exhibits to angular acceleration when a torque is applied. Imagine trying to spin two objects: a solid disk and a hollow ring of the same mass and radius. You'll find that the hollow ring is harder to spin. This happens because mass distribution affects the moment of inertia.In the case of a solid cylinder, the moment of inertia around its central axis is calculated with a simple formula: \[I_{\text{cm}} = \frac{1}{2} M R^2\]Here, \(M\) is the mass and \(R\) is the radius of the cylinder. Understanding this concept helps in determining how objects will react under different rotational forces, making it crucial for engineering and physics applications.
Solid Cylinder
A solid cylinder is a three-dimensional shape with a fixed radius and height. It is defined by its circular base and consistent cross-section. In physics, solid cylinders often serve as basic subjects in experiments and calculations because of their simple shape.The solid cylinder's mass distribution is uniform, meaning the mass is spread evenly across its entire volume. This uniformity simplifies calculations such as finding the mass \(M\) or using the moment of inertia formula.Understanding a solid cylinder's characteristics is pivotal in using the parallel axis theorem, which helps calculate its moment of inertia about different axes.
Center of Mass
The center of mass, often denoted as CoM, is the point where the entire mass of an object is thought to be concentrated. It acts as the balance point of the object. For regular shapes like solid cylinders, the center of mass lies exactly at the geometric center. In calculations, the center of mass is key because it simplifies complex systems to simpler, solvable problems. For a solid cylinder, finding the moment of inertia about any axis involves understanding where the center of mass lies. Knowing that the CoM is at the center allows us to easily apply the parallel axis theorem to find moment of inertia for different rotating axes.
Physics Education
Physics education often relies on fundamental principles to help students grasp complex ideas. A core tool used in teaching physics is breaking down problems into simpler solutions, such as understanding the moment of inertia or using the parallel axis theorem. A solid cylinder provides a straightforward example to make these concepts clear. This educational approach involves:
  • Applying real-life examples to physics problems.
  • Using visual aids to illustrate abstract ideas.
  • Breaking problems into smaller, more digestible parts.
By understanding basic shapes like cylinders and using the parallel axis theorem, students learn how to solve more complex problems efficiently. This hands-on problem-solving strategy ensures that students can apply learned concepts to real-world scenarios.

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Most popular questions from this chapter

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of 27.0 above the horizontal. One child, with a mass of 44.0 kg, is hanging 1.30 m along the branch from the tree trunk. The other child, with a mass of 35.0 kg, is hanging 2.10 m from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.

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 s. \((\text { a) Find the magnitude of the angular deceleration of the }\) cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

A uniform board is leaning against a smooth vertical wall. The board is at an angle above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is 0.650. Find the smallest value for the angle , such that the lower end of the board does not slide along the ground.

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.

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