Question

vi) calculate moment of inertia about an axis pasid through centre of mass of the disc of radius. and having moss 5 k 5 kg.​

Answer 1

Answer:

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Explanation:

objects

In the preceding section, we defined the moment of inertia but did not show how to calculate it. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. This section is very useful for seeing how to apply a general equation to complex objects (a skill that is critical for more advanced physics and engineering courses).

Moment of Inertia

We defined the moment of inertia I of an object to be

\[I=\sum _{i}{m}_{i}{r}_{i}^{2}\]

for all the point masses that make up the object. Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let’s take a simple example of two masses at the end of a massless (negligibly small mass) rod ((Figure)) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.

In the case with the axis in the center of the barbell, each of the two masses m is a distance R away from the axis, giving a moment of inertia of

\[{I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}.\]

In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is

\[{I}_{2}=m{(0)}^{2}+m{(2R)}^{2}=4m{R}^{2}.\]

From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center.

Answer 2

Explanation:

The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. (Figure) shows a point P as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

A lamina is perfectly balanced on a spindle if the lamina’s center of mass sits on the spindle.