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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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Most popular questions from this chapter

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 .

\({ }^{\dagger} \mathrm{A}\) uniform solid sphere of radius \(a\) rolls without slipping inside a fixed sphere of radius \(2 a\). Show that if \(e\) is the unit vector pointing from the centre of the larger sphere towards the centre of the smaller sphere, then $$ 7 a e \wedge \ddot{e}-2 a n \dot{e}+5 g e \wedge k=0 $$ where \(n\) is constant and \(k\) is a unit vector in the direction of the upward vertical. The dot denotes the time derivative with respect to fixed axes.

Show that the inertia matrix at the centre of a uniform solid cube with mass \(m\) and edges of length \(2 a\) is \(\frac{2}{3} m a^{2} I\). Find the principal axes and principal moments of inertia at a vertex.

\({ }^{\dagger} \mathrm{A}\) thin uniform rod of length \(2 a\) and mass \(m\) has a small light ring fixed at one end. The ring is threaded on a fixed vertical wire. Show that if \(z\) is the height of the centre of the rod, \(\theta\) the angle the rod makes with the upward vertical, and \(\varphi\) the angle that the vertical plane containing the rod makes with a fixed vertical plane, then the Lagrangian of the system is $$ L=\frac{1}{6} m\left[3 \dot{z}^{2}+a^{2} \dot{\theta}^{2}\left(1+3 \cos ^{2} \theta\right)+4 a^{2} \dot{\varphi}^{2} \sin ^{2} \theta\right]-m g z . $$ Initially the rod makes an acute angle \(\alpha\) with the vertical and its centre has velocity \(V\) perpendicular to the rod and the wire. Show that the angle the rod makes with the wire oscillates between \(\alpha\) and \(\pi-\alpha\) with period $$ \frac{a}{V} \int_{-\cos \alpha}^{\cos \alpha}\left[\frac{1+3 u^{2}}{\cos ^{2} \alpha-u^{2}}\right]^{1 / 2} \mathrm{~d} u $$

\({ }^{\dagger} \mathrm{A}\) thin uniform disc, mass \(M\), radius \(a\) and centre \(C\), has a thin uniform rod \(O C\), mass \(m\) and length \(a \sqrt{3}\), fixed to it at \(C\), so that \(O C\) is orthogonal to the disc. The end \(O\) of the rod is fixed but freely pivoted at the centre \(O\) of a horizontal turntable, and the rim of the disc rests on the surface of the turntable. No slipping occurs. The turntable is forced to rotate about the vertical axis through \(O\) with variable angular velocity \(\Omega\). Initially the system is at rest. Show that if \(P\) is the point of contact between the disc and the turntable, and \(\varphi\) is the angle between \(O P\) and a line fixed in the turntable, then $$ \dot{\varphi}=-\left(\frac{11 M+4 m}{19 M+4 m}\right) \Omega $$
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