In: Chemistry

Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length (I=M ℓ² / 3), prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is I=M ℓ² / 12. You will find the graphics in Figure 10.12 useful in visualizing these rotations.

A moment of inertia represents the body's tendency to resist rotational motion. It is given by the product of mass and square of the distance from the axis of rotation.

I=M R²

The parallel axis theorem states that: the moment of inertia increases when the axis of rotation moves away. It is the sum of moment of inertia through the center of mass and product of group and square of perpendicular distance from the rotation axis and center of the assembly.

I=Ic m+M d²

We know that when the rod rotates about a parallel axis passing through the end of the rod, the moment of inertia is:

I=1/3 M l²

Where M is the mass of the rod and l is the length of the rod.

The distance from the end of the rod to the center is d=l/2.

I=Ic m+M d²

=> Icm=I-M d²

Therefore:

Icm =1/3 Ml²-M(l/2)²

Icm =1/3 Ml²-1/4 Ml²

Icm =1/12 Ml²

Hence proved that moment of inertia of the rod about the axis passing through its center of mass and perpendicular to it is Ml²/12

Two balls connected by a rod, as shown in the figure below (Ignore rod’s mass). Mass of ball X is 8kg and the mass of ball Y is 5 kg What is the moment of inertia of the system about AB?

To calculate moment of inertia of axis passing through centre of a square

Two balls connected by a rod as shown in the figure below (Ignore rod’s mass). What is the moment of inertia of the system?
Given :
mX = 400 grams = 0.4kg
mY = 500 grams = 0.5 kg
rX = 0cm = 0m
rY = 30cm = 0.3m

Three rods each of mass m and length I are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of triangle.

explain how to calculate the moment of inertia of a disk, we will take the example of a uniform thin disk which is rotating about an axis through its centre.

From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as . Calculate the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc.

From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as . Calculate the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc.

On which of the following does the moment of inertia of an object depend?
Check all that apply.
linear speed
linear acceleration
angular speed
angular acceleration
total mass
shape and density of the object
location of the axis of rotation

Find the moment of inertia Ix of particle a with respect to the x-axis (that is, if the x-axis is the axis of rotation), the moment of inertia Iy of particle a with respect to the y axis, and the moment of inertia Iz of particle a with respect to the z-axis (the axis that passes through the origin perpendicular to both the x and y axes).
Express your answers in terms of m and r separated by commas.

A toy top with a spool of diameter 5.0cm has a moment of inertia of 3.0x10^-5kg x m^2 about its rotation axis. To get the top spinning, its string is pulled with a tension of .30 N. How long does it take for the top to complete the first five revolutions? The string is long enough that it is wrapped around the top more than five turns