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Chapter 5: Problem 3

\(^{\dagger}\) Show that the principal moments of inertia at the centre of mass of a uniform solid circular cylinder, radius \(a\), height \(2 h\), and mass \(m\), are \(\frac{1}{2} m a^{2}\) and \(\frac{1}{12} m\left(4 h^{2}+3 a^{2}\right)\) (repeated). Find the principal axes and principal moments of inertia at a point distance \(D\) from the centre of mass in the plane through the centre of mass perpendicular to the axis of the cylinder.

Short Answer

Expert verified
Answer: The new principal moments of inertia at point A are \(I'_1 = I'_x = \frac{1}{12} m(4h^2 + 3a^2) + mD^2\), \(I'_2 = I'_y = \frac{1}{12} m(4h^2 + 3a^2) + mD^2\), and \(I'_3 = I'_z = \frac{1}{2} m a^2\). The new principal axes will be the same as the original principal axes since the cylinder is rotationally symmetric about the z-axis.

Step by step solution

01

Find the moments of inertia of the circular cylinder at its center of mass

For a uniform solid circular cylinder with radius a, height 2h, and mass m, we have \(I_z=\frac{1}{2}ma^2\) and \(I_x=I_y=\frac{1}{12}m(4h^2 + 3a^2)\) The moments of inertia are calculated using standard formulas for a uniform solid circular cylinder, considering its rotational symmetry about the z-axis.
02

Find the principal moments of inertia and axes

The principal moments of inertia at the center of mass are the moments of inertia along the principal axes. Since the cylinder is rotationally symmetric about the z-axis, its moments of inertia about the x- and y-axes are the same. Therefore, the principal moments of inertia at the center of mass are: \(I_1 = I_x = \frac{1}{12} m(4h^2 + 3a^2)\) \(I_2 = I_y = \frac{1}{12} m(4h^2 + 3a^2)\) \(I_3 = I_z = \frac{1}{2} m a^2\) Thus, the repeated moment of inertia is \(\frac{1}{12} m(4h^2 + 3a^2)\).
03

Find the moments of inertia at a point distance D from the center of mass

To find the moments of inertia at a point distance D from the center of mass, we use the parallel axis theorem. Let A be the new point distance D from the center of mass in the plane through the center of mass perpendicular to the axis of the cylinder. \(I'_x = I_x + mD^2\) \(I'_y = I_y + mD^2\) \(I'_z = I_z\)
04

Find the new principal moments of inertia and axes

At point A, the new moments of inertia are: \(I'_1 = I'_x = \frac{1}{12} m(4h^2 + 3a^2) + mD^2\) \(I'_2 = I'_y = \frac{1}{12} m(4h^2 + 3a^2) + mD^2\) \(I'_3 = I'_z = \frac{1}{2} m a^2\) So, the principal moments of inertia at point A are \(\frac{1}{2} m a^{2}\) (non-repeated) and \(\frac{1}{12} m\left(4 h^{2}+3 a^{2}\right) + mD^2\) (repeated). The new principal axes will be the same as the original principal axes since the cylinder is rotationally symmetric about the z-axis, although the moments of inertia have changed.

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Key Concepts

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

Circular Cylinder
A circular cylinder is a 3D geometric shape that is extremely important in physics, especially in mechanics and fluid dynamics. When we talk about a circular cylinder in the context of rotational dynamics, we usually consider its mass, radius, and height.
The form resembles a soda can, where the top and bottom are circles of equal radius, denoted as \( a \), and the distance between these bases is the height, often labeled as \( 2h \) in mathematical problems.
This shape is quite symmetrical and serves as an excellent example for studying rotational and inertia-related concepts. The mass \( m \) of the cylinder distributes evenly across its volume, which is essential for calculating its moments of inertia. The mass distribution impacts how the cylinder rotates about different axes.
Principal Axes
Principal axes are specific directions in an object through which the moments of inertia are either minimized or maximized. In simpler terms, they are the axes about which an object rotates most naturally without any complex motion.
For a circular cylinder, which is rotationally symmetric, the principal axes align with the x, y, and z-axes.
  • The axis along the height of the cylinder is usually the z-axis.
  • The other two axes, the x and y, lie in the plane that crosses the center of mass perpendicular to the z-axis.
Along these axes, the cylinder experiences consistent resistance when rotated, making these axes very special for analyzing rotational motion. Understanding these axes allows physicists to simplify complex rotational systems.
Parallel Axis Theorem
The parallel axis theorem is a handy tool in physics for simplifying calculations of rotational inertia when the axis of rotation shifts away from the center of mass of a body.
This theorem says that if you know the moment of inertia of a body about an axis through its center of mass, you can find the moment about any parallel axis. You simply add \( mD^2 \) to the original moment, where \( D \) is the distance between the axes.
Thus, for any solid object like our cylinder, this theorem helps us extend our moment of inertia calculations to axes that are not quite as straightforward or conveniently located.
Rotational Symmetry
Rotational symmetry is a property that allows an object to look the same after a certain amount of rotation. For our circular cylinder, rotational symmetry is observed about its central longitudinal axis.
This symmetry is significant because it implies that the cylinder's properties, specifically moments of inertia, remain unchanged when rotated around the z-axis.
Consequently, it simplifies calculations as the principal moments of inertia about the x and y-axes are equal due to this symmetry, reducing the complexity in analyzing and designing systems involving cylindrical rotation.
The presence of rotational symmetry often simplifies mathematical models and equations necessary for understanding the distribution and effects of rotational dynamics.

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Most popular questions from this chapter

The surface of a rigid body is an ellipsoid with equation $$ A x^{2}+B y^{2}+C z^{2}=k^{2} $$ in a rest frame with origin \(O\) at the centre of mass and axes aligned with the principal axes at the centre of mass; \(A, B\), and \(C\) are the principal moments of inertia at \(O\) and \(k\) is a constant. Show that if the body rotates freely about \(O\), with \(O\) fixed relative to an inertial frame, then the two tangent planes to the surface at the intersection points between the surface and the instantaneous axis are fixed relative to the inertial frame. Deduce that the body moves as if it were rolling between these two planes.

A top of mass \(m\) is pivoted at a point on its axis of symmetry at a distance \(a\) from its centre of mass. The three principal moments of inertia at the pivot are \(A, A, C\), where \(C\) is the moment of inertia about the symmetry axis. Show that, with an appropriate choice of Euler angles, the Lagrangian is $$ L=\frac{1}{2} A\left(\dot{\theta}^{2}+\dot{\varphi}^{2} \sin ^{2} \theta\right)+\frac{1}{2} C(\dot{\psi}+\dot{\varphi} \cos \theta)^{2}-m g a \cos \theta . $$ Show that \(n=\dot{\psi}+\dot{\varphi} \cos \theta\) and \(k=A \dot{\varphi} \sin ^{2} \theta+C n \cos \theta\) are constants of the motion. Write down the total energy and explain briefly why it is also a constant of the motion. The top is set in motion with $$ \theta=\frac{\pi}{3}, \quad \dot{\theta}=0, \quad \dot{\varphi}=\frac{2 \gamma}{\sqrt{3}}, \quad \dot{\psi}=\frac{(3 A-C) \gamma}{C \sqrt{3}} $$ where \(\gamma=\sqrt{m g a / A}\). Show that $$ \left(\frac{\mathrm{d} u}{\mathrm{~d} t}\right)^{2}=\gamma^{2}(1-u)^{2}(2 u-1) $$ where \(u=\cos \theta\). Verify that $$ \frac{1}{u}=1+\frac{1}{\cosh (\gamma t)} \text {. } $$ What happens to the axis of the top as \(t \rightarrow \infty\) ? See [13], p. 158 .

Kovalevskaya's top has Lagrangian $$ L=C\left(\dot{\theta}^{2}+\dot{\varphi}^{2} \sin ^{2} \theta+\frac{1}{2}(\dot{\psi}+\dot{\varphi} \cos \theta)^{2}\right)+m g a \sin \theta \cos \psi $$ where \(C\) and \(m\) are constants. Describe the physical system that has this Lagrangian. Note that \(L\) is independent of \(\varphi\) and \(t\) and write down the corresponding conserved quantities. Put $$ z=C(\dot{\varphi} \sin \theta+\mathrm{i} \dot{\theta})^{2}+m g a \sin \theta \mathrm{e}^{-\mathrm{i} \psi} $$ Show that $$ \frac{\mathrm{d} z}{\mathrm{~d} t}=\mathrm{i}(\dot{\varphi} \cos \theta-\dot{\psi}) z $$ and deduce that \(|z|^{2}\) is also conserved. See [13], p. 166 .

\({ }^{\dagger} \mathrm{A}\) hollow right circular cylinder is fixed with its axis vertical. A uniform solid sphere rolls without slipping on the inside surface of the cylinder. Show that when the centre of the sphere does not move on a vertical line, the height of the centre of the sphere performs simple harmonic motion, and that between oscillations the plane containing the axis of the cylinder and the centre of the sphere turns through an angle \(\pi \sqrt{14}\).

Show that the inertia matrix at the centre of any uniform Platonic solid is a multiple of the identity matrix.
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