Skip to main content
Ch 09: Rotation of Rigid Bodies
Chapter 9, Problem 9

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis (d) parallel to the bar and 0.500 m from it.

Verified step by step guidance
1
Identify the components of the system and their respective masses and distances from the axis of rotation. Here, the bar has a mass of 4.00 kg and the balls each have a mass of 0.300 kg. The total length of the bar is 2.00 m, so each ball is 1.00 m from the center of the bar.
Calculate the moment of inertia of the bar about its center using the formula for a rod rotating about its center: $I_{ ext{bar}} = \frac{1}{12} M L^2$, where $M$ is the mass of the bar and $L$ is its length.
Use the parallel axis theorem to find the moment of inertia of the bar about the new axis, which is 0.500 m from the center. The parallel axis theorem states $I = I_{ ext{cm}} + Md^2$, where $I_{ ext{cm}}$ is the moment of inertia about the center of mass, $M$ is the mass, and $d$ is the distance between the two axes.
Calculate the moment of inertia of each ball about the new axis. Since the balls can be treated as point masses, use the formula $I = mr^2$, where $m$ is the mass of the ball and $r$ is the distance from the axis to the ball. The distance from the new axis to each ball is 0.500 m plus or minus the distance from the center of the bar to the ball.
Sum the moments of inertia of the bar and the two balls to find the total moment of inertia of the system about the specified axis.

Verified Solution

Video duration:
6m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to rotational motion about a specific axis. It depends on the mass distribution relative to that axis. For point masses, it is calculated as the product of the mass and the square of the distance from the axis of rotation. The greater the distance or mass, the higher the moment of inertia.
Recommended video:
Guided course
11:47
Intro to Moment of Inertia

Parallel Axis Theorem

The parallel axis theorem allows us to calculate the moment of inertia of a body about any axis parallel to an axis through its center of mass. It states that the moment of inertia about the new axis is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes. This theorem is essential for solving problems involving composite bodies.
Recommended video:
Guided course
13:46
Parallel Axis Theorem

Center of Mass

The center of mass is the point at which the mass of a system is concentrated and can be considered to act. For a uniform bar with point masses at its ends, the center of mass can be found by considering the distribution of mass along the bar. Understanding the center of mass is crucial for calculating the moment of inertia, as it serves as a reference point for applying the parallel axis theorem.
Recommended video:
Guided course
06:30
Intro to Center of Mass
Related Practice
Textbook Question
A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis (b) perpendicular to the bar through one of the balls;
520
views
Textbook Question
A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis (c) parallel to the bar through both balls;
549
views
Textbook Question
A compound disk of outside diameter 140.0 cm is made up of a uniform solid disk of radius 50.0 cm and area density 3.00 g/cm^2 surrounded by a concentric ring of inner radius 50.0 cm, outer radius 70.0 cm, and area density 2.00 g^cm^2. Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.
933
views
Textbook Question
A thin, rectangular sheet of metal has mass M and sides of length a and b. Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.
1723
views
Textbook Question
Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.
1064
views
Textbook Question
Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits 10^-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. (b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
1130
views