**Moment of Inertia**Derive the formula for the moment of inertia of a uniform thin rod of length \( L \) and mass \( M \) about an axis through its center, perpendicular to its face.*Diagram Explanation:* The diagram shows a horizontal rod with a vertical dashed line through its center, indicating the axis of rotation.Repeat the calculation, only now assume the rod has a density that increases uniformly from a value of \( \rho_0 \) on one end to \( 2\rho_0 \) on the other end.**Suggestion**For the second integration, it is important to define the density function correctly: It will be a straight line that includes the two points \((-L/2, \rho_0)\) and \((L/2, 2\rho_0)\). Find the slope of the line as rise over run, and the x-intercept, \( \rho(0) \).Total mass of the rod, \( M \), will be the integral of the density function from \(-L/2\) to \( L/2 \).If you solve the second integration correctly, including the value of the total mass, \( M \), in terms of \( \rho \), you should see a very interesting relationship between the two moments of inertia.

Given: The mass of the rod is M. The length of the rod is L.

Answered: Moment of inertia Derive the formula…

Moment of inertia Derive the formula for the moment of inertia of a uniform thin rod of length L and mass M about an axis through its center, perpendicular to its face.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: The moment of inertia of a solid sphere about an axis passing through the centre of gravity is 2/5MR…

A: By parallel axis theorem, the moment of inertia at 2R is I = (2MR2/5) + M(2R)2 I = 22MR2/5

Q: Barber's Problem: Calculate the moment of inertia of the object made of a thin hoop of mass M and…

Q: What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through…

A: The moment of inertia of rod about an axis at one end and perpendicular to it is:

Q: A uniform flat disk of radius R and mass 2M is pivoted at point P. A point mass of 1/2 M is attached…

A: Formula used:Moment of inertia;I = (1/2)mr2The parallel axis theorem:I = Ic +md2WHERE:Ic = moment of…

Q: A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.78 m and a…

Q: Move over Jeff Bezos, this invention is about to make you the richest person in the world. You have…

Q: Consider two uniform solid spheres where both have the smae diameter but one has twice the mass of…

Q: what torque is reuired to uniformly rotate a disk with moment of inertia 0.7kgm^2 up to 2 rad/s in 5…

Q: Calculate the moment of inertia by direct integration of a thin rod of mass M and length L about an…

Q: 21

Q: The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender…

Q: xis through its center. When the skater’s hands and arms are brought in and wrapped around his body…

A: Given:- His hands and arms have a combined mass of m = 8.0 kg When outstretched, they span 1.8 m…

Q: A CD has a mass of 17 g and a radius of 6.0cm. When inserted into a player, the CD starts from rest…

A: Given information: The mass of the CD (m) = 17 g = 0.017 kg The radius of the CD (r) = 6.0 cm = 0.06…

Q: Calculate the rotational inertia of a long, thin rod of length L and mass M about one end. Assume…

A: Rotational inertia about on end of the rod I=∫0Lx2dm=∫0Lx2λdx=∫0Lx2λ01+xLdx

Q: A uniform rod of length L and mass M is held vertically with one end resting on the floor as shown…

Q: Consider two uniform solid spheres where one has twice the mass and twice the diameter of the other.…

A: The moment of inertia of the solid sphere is given as I=25MR2 Here, M is the mass and R is the…

Q: Calculate the moment of inertia of a hollow sphere of mass M and radius R about its diameter.

A: In order to determine the moment of inertia of hollow sphere (mass=M, radius=R) about its diameter,…

Q: A dragster starts from rest and accelerates down a track. Each tire has a radius of 0.320 m and…

Q: A rod of mass 3m and length L can pivot about an axis through its center and perpendicular to the…

Q: The figure below shows a rigid structure consisting of a circular hoop of radius R and mass m, and a…

A: Given: T=6.6 s R=0.60 m m=6.0 kg

Question

**Moment of Inertia**

Derive the formula for the moment of inertia of a uniform thin rod of length \( L \) and mass \( M \) about an axis through its center, perpendicular to its face.

*Diagram Explanation:*
The diagram shows a horizontal rod with a vertical dashed line through its center, indicating the axis of rotation.

Repeat the calculation, only now assume the rod has a density that increases uniformly from a value of \( \rho_0 \) on one end to \( 2\rho_0 \) on the other end.

**Suggestion**

For the second integration, it is important to define the density function correctly: It will be a straight line that includes the two points \((-L/2, \rho_0)\) and \((L/2, 2\rho_0)\). Find the slope of the line as rise over run, and the x-intercept, \( \rho(0) \).

Total mass of the rod, \( M \), will be the integral of the density function from \(-L/2\) to \( L/2 \).

If you solve the second integration correctly, including the value of the total mass, \( M \), in terms of \( \rho \), you should see a very interesting relationship between the two moments of inertia.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

Calculate moment of inertia of a desk with a mass of 6.00 kg and a radius of 0.250 m, with a axis of rotational half way between its center and it’s rim.
The uniform thin rod in the figure below has mass M = 2.00 kg and length L = 3.25 m and is free to rotate on a frictionless pin. At the instant the rod is released from rest in the horizontal position, find the magnitude of the rod's angular acceleration, the tangential acceleration of the rod's center of mass, and the tangential acceleration of the rod's free end.
A solid cylinder of uniform density of radius 2 cm has mass of 50 g. If its length is 10 cm, Calculate its moment of inertia about (i) its own axis of rotation passing through the centre. (ii) an axis passing through its centre and perpendicular to its length.
The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Figure 1). When his hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass 8.0 kg. When outstretched, they span 1.6 m ; when wrapped, they form a thin-walled hollow cylinder of radius 26 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.4 kg⋅m2. If his original angular speed is 0.30 rev/s, what is his final angular speed?
Determine the moment of inertia of a solid homogeneous cylinder of radius R and length L with respect to a diameter in the base of the cylinder.
Find the moment of inertia for a uniform rod (mass m and length L) about an axis at the position shown in the figure
A uniform, solid cylinder with mass M and radius 2R rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass M and radius R that is mounted on a frictionless axle through its center. A block of mass M is suspended from the free end of the string (the figure (Figure 1)). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest. Express your answer in terms of g, M, R.
Find the moment of inertia for a solid cube is bounded by planes x = F1, z = 71 , y = 3 and y = 5 about x – axis