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Chapter 10: Problem 1

Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is $$ I_{c}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be $$ I=I_{c}+M\left(\frac{L}{2}\right)^{2}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}+\frac{L^{2}}{4}\right)=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{3}\right) $$ When \(L=6 R, \quad I_{h}=\frac{49}{4} M R^{2}\)

Short Answer

Expert verified
The moment of inertia is \( I = \frac{49}{4} M R^2 \).

Step by step solution

01

Identify the Given Equation

The problem provides the moment of inertia of a cylinder around an axis through its center and perpendicular to its axis as \( I_c = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) \). We need to use this to find the moment of inertia around the edge using the parallel axis theorem.
02

Apply the Parallel Axis Theorem

The parallel axis theorem states that if you know the moment of inertia about an axis through the center of mass, the moment of inertia about any parallel axis is \( I = I_c + Md^2 \), where \( d \) is the perpendicular distance between the axes. For this problem, \( d = \frac{L}{2} \).
03

Calculate Moment of Inertia Using Theorem

Substitute \( I_c = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) \) and \( d = \frac{L}{2} \) into the parallel axis theorem formula: \[ I = I_c + M\left(\frac{L}{2}\right)^2 = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) + M\left(\frac{L^2}{4}\right) \].
04

Simplify the Expression

Combine the terms inside the brackets:\[ I = M\left(\frac{R^2}{4} + \frac{L^2}{12} + \frac{L^2}{4}\right) \].Simplify further to get:\[ I = M\left(\frac{R^2}{4} + \frac{3L^2}{12}\right) = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right) \].
05

Substitute Given Values

Using the condition \( L = 6R \), substitute into the simplified expression:\[ I = M\left(\frac{R^2}{4} + \frac{(6R)^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{36R^2}{3}\right) \].
06

Final Calculation

Simplify the terms:\[ I = M\left(\frac{R^2}{4} + 12R^2\right) = M\left(\frac{R^2}{4} + \frac{48R^2}{4}\right) = M\left(\frac{49R^2}{4}\right) \],which gives:\[ I = \frac{49}{4} MR^2 \].

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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 key principle when studying rotational dynamics and calculating the moment of inertia around different axes. It allows you to find the moment of inertia about any axis that is parallel to an axis through the center of mass.

Here's how it works:
  • It starts with the known moment of inertia about the center of mass, referred to as Ic.
  • The formula is: \( I = I_c + Md^2 \), where:
    • \( I \) is the moment of inertia about the new axis.
    • \( M \) is the mass of the object.
    • \( d \) is the distance between the two axes.
For a cylinder, if you have its moment of inertia about the center of mass, you can easily shift it along a parallel axis.

This theorem is widely used in engineering and physics to simplify complex calculations by breaking them into simpler parts.
Cylinder
A cylinder is one of the most common shapes considered in rotational motion and mechanics. Understanding its geometry and properties is essential to solving problems related to its moment of inertia.

Here's what makes it unique:
  • A cylinder has a circular base, radius \( R \), and height \( L \).
  • Moments of inertia are typically calculated about axes that are centered and perpendicular to its base.
  • For a solid cylinder, the moment of inertia about its central axis is given by \( I_c = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \).
The symmetry and regularity of a cylinder simplify many calculations, especially when using principles like the parallel axis theorem. Cylinders often serve as basic models for more complex, real-world objects.
Rotational Motion
Rotational motion describes objects that spin around an axis. Unlike linear motion, rotational motion involves angular velocity, angular acceleration, and moments of inertia.

Key points about rotational motion include:
  • It is characterized by the axis of rotation, which can be at a point within the object or outside it.
  • Moment of inertia is crucial as it defines how difficult it is to change the object's rotation.
  • Angular velocity (\( \omega \)) is the rate of rotation and is analogous to linear velocity.
Understanding these concepts helps explain how objects like wheels, discs, and even planets move through space. Calculating the moment of inertia in various scenarios, such as using the parallel axis theorem, provides deeper insight into these motions.

By analyzing rotational motion, one can gain greater control and predictability over engineered systems and natural phenomena alike.

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Most popular questions from this chapter

The period of revolution of planet \(A\) around the sun is 8 times that \(B\). The distance of a from the sun is how many times greater than that of \(B\) from the sun? (a) 2 (b) 3 (c) 4 (d) 5

A space ship moves from earth to moon and back. Th. greatest energy required for the space ship is \(t\) overcome the difficulty in (a) entering the earth's gravitational field (b) take off from earth's field (c) take off from lunar surface (d) entering the moon's lunar surface

If satellite is revolving around a planet of mass \(M\) in an elliptical orbit of semi-major axis \(a\), find the orbital speed of the satellite when it is at a distance \(r\) from the focus. (a) \(v^{2}=G M\left[\frac{2}{r}-\frac{1}{a}\right]\) (b) \(v^{2}=G M\left[\frac{2}{r^{2}}-\frac{1}{a}\right]\) (c) \(v^{2}=G M\left[\frac{2}{r^{2}}-\frac{1}{a^{2}}\right]\) (d) \(v^{2}=G\left[\frac{2}{r}-\frac{1}{a}\right]\)

Mass of moon is \(7.34 \times 10^{22} \mathrm{~kg}\). If the acceleration due to gravity on the moon is \(1.4 \mathrm{~ms}^{-2}\), the radius of the moon is \(\left(G=6.667 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right)\) (a) \(0.56 \times 10^{4} \mathrm{~m}\) (b) \(1.87 \times 10^{6} \mathrm{~m}\) (c) \(1.92 \times 10^{6} \mathrm{~m}\) (d) \(1.01 \times 10^{8} \mathrm{~m}\)

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon \(\quad\) [NCERT Exemplar] (a) will be elliptical (b) will not be strictly elliptical because the total gravitational force on it is not central (c) is not elliptical but will necessarily be a closed curve (d) deviates considerably from being elliptical due to influence of planets other than earth
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