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(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be \(2 M R^{2} / 5\), where \(M\) is the mass of the sphere and \(R\) is the radius of the sphere. (b) Given the moment of inertia of a disc of mass \(M\) and radius \(R\) about any of its diameters to be \(M R^{2} / 4\), find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Short Answer

Expert verified
The moment of inertia of the sphere about a tangent is \( \frac{7}{5} M R^2 \) and for the disc about the edge axis is \( \frac{3}{2} M R^2 \).

Step by step solution

01

Identify Known Values for Sphere

We know the moment of inertia of the sphere about any of its diameters: \( I_d = \frac{2}{5} M R^2 \). Our goal is to find the moment of inertia about a tangent.
02

Apply Parallel Axis Theorem for Sphere

The parallel axis theorem states: \( I = I_d + Md^2 \) where \( d \) is the distance from the axis through the center to the parallel axis (the tangent). For a sphere, \( d = R \), so \( I = \frac{2}{5} M R^2 + M R^2 \).
03

Calculate Moment of Inertia for Tangent Axis

Substitute into the equation: \[ I = \frac{2}{5} M R^2 + M R^2 = \frac{2}{5} M R^2 + \frac{5}{5} M R^2 = \frac{7}{5} M R^2. \] Thus, the moment of inertia of the sphere about a tangent is \( \frac{7}{5} M R^2 \).
04

Identify Known Values for Disc

We know the moment of inertia of the disc about any diameter: \( I_d = \frac{1}{4} M R^2 \). We need the moment of inertia about an axis normal to the disc and on the edge.
05

Apply Parallel Axis Theorem for Disc

The parallel axis theorem is used again, now for the disc. The axis through the center and perpendicular to the plane of the disc has \( I_c = \frac{1}{2} M R^2 \). The axis of interest is at distance \( R \) from this central axis.
06

Calculate Moment of Inertia for Edge Axis

Using the parallel axis theorem: \[ I = I_c + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2. \] Thus, the moment of inertia about the edge is \( \frac{3}{2} M R^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel Axis Theorem
The Parallel Axis Theorem is a very helpful tool in physics when calculating the moment of inertia for a body about an axis that is parallel to one passing through its center of mass. It allows us to easily adjust the known moment of inertia through the central axis to find it about another axis.

According to the theorem, if you know the moment of inertia of a body about an axis that passes through its center of mass, noted as \( I_c \), you can find the moment of inertia about another axis that is parallel to this and a distance \( d \) away using this formula:

\[ I = I_c + Md^2 \]

Here, \( M \) is the mass of the body and \( d \) is the perpendicular distance between the two axes.
  • It requires accurate knowledge of the distance \( d \).
  • This theorem simplifies situations without having to individually recalculate the total inertia, saving both time and effort.
This theorem is frequently used in problems involving bodies like spheres or discs.
Sphere
Spheres are symmetrical three-dimensional shapes, and understanding their physical properties is important in mechanics. Specifically, when dealing with calculations of a sphere's moment of inertia, the symmetry simplifies things greatly.

The moment of inertia of a solid sphere about any of its diameters is given by \( \frac{2}{5} M R^2 \), where \( M \) is the mass and \( R \) is the radius. Using this knowledge, along with the parallel axis theorem, we can determine the moment of inertia about various other axes, such as a tangent to the sphere.
  • Perfect symmetry reduces complexity in calculations.
  • Knowing the diameter inertia helps calculate inertia about any axis through it.
This fundamental understanding of a sphere's inertia helps in various practical applications, from computing rotations to engineering problems.
Disc
A disc is another key shape often found in rotational mechanics. It's important to note that a disc's inertia properties can vary depending on the axis.

A crucial point to know is that the moment of inertia of a disc about a diameter is \( \frac{1}{4} M R^2 \). But when we consider an axis perpendicular to the disc's plane (like a central axis running vertically), the moment of inertia becomes \( \frac{1}{2} M R^2 \).
  • The central axis inertia is greater than the diameter inertia.
  • This affects how the disc responds to forces and torques applied externally.
When dealing with a disc, these variables and relations must always be kept in mind for accurate calculations.
Tangent Axis
A tangent axis refers to an imaginary line that touches the edge of a body, like a circle or sphere, but doesn’t pass through its center. Calculating the moment of inertia about such an axis can initially seem more challenging.

The parallel axis theorem is invaluable here. It allows us to transition the measure from a known axis to a tangent axis. For example, with a sphere, if you know its moment of inertia about the diameter, you can calculate it about a tangent by adjusting for the radius, as shown in the solution with:
\[ I = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 \]

This illustrates how symmetry and the theorem work together to make even complex calculations manageable.
Normal Axis
A normal axis is an axis that extends perpendicular to the plane of the object, like our disc example. Understanding motion about a normal axis is crucial for comprehending rotational dynamics.

In practical applications, like the one in the exercise, knowing the moment of inertia about an axis perpendicular to the disc's core (central axis), we see that it equals \( \frac{1}{2} M R^2 \).
  • This contrasts with in-plane axes and is generally more significant due to the distance from the center.
  • Understanding this helps in forming accurate predictions on how an object will behave under rotational forces.
This, coupled with the Parallel Axis Theorem, allows us to understand the inertia about other important axes, like the edge.

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Most popular questions from this chapter

From a uniform disk of radius \(R\), a circular hole of radius \(R / 2\) is cut out. The centre of the hole is at \(R / 2\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

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