The axis of a gyroscope is free to rotate within a smooth horizontal circle in colatitude \(\lambda\). Due to the Coriolis force, there is a couple on the gyroscope. To find the effect of this couple, use the equation for the rate of change of angular momentum in a frame rotating with the Earth (e.g., that of Fig. 5.7), \(\dot{\boldsymbol{J}}+\boldsymbol{\Omega} \wedge \boldsymbol{J}=\boldsymbol{G}\), where \(\boldsymbol{G}\) is the couple restraining the axis from leaving the horizontal plane, and \(\boldsymbol{\Omega}\) is the Earth's angular velocity. (Neglect terms of order \(\Omega^{2}\), in particular the contribution of \(\boldsymbol{\Omega}\) to \(\boldsymbol{J} .\) ) From the component along the axis, show that the angular velocity \(\omega\) about the axis is constant; from the vertical component show that the angle \(\varphi\) between the axis and east obeys the equation \(I_{1} \ddot{\varphi}-I_{3} \omega \Omega \sin \lambda \cos \varphi=0\) Show that the stable position is with the axis pointing north. Determine the period of small oscillations about this direction if the gyroscope is a flat circular disc spinning at 6000 r.p.m. in latitude \(30^{\circ} \mathrm{N}\). Explain why this system is sensitive to the horizontal component of \(\boldsymbol{\Omega}\), and describe the effect qualitatively from the point of view of an inertial observer.

Step by step solution

01

Write down the given equation

The rate of change of angular momentum equation is given by: \(\dot{\boldsymbol{J}}+\boldsymbol{\Omega} \wedge \boldsymbol{J}=\boldsymbol{G}\)

02

Obtain the component along the axis

To show that the angular velocity \(\omega\) about the axis is constant, we need to find the component of the equation along the axis. In order to do this we can take the dot product of the given equation with the unit vector along the axis, \(\hat{z}\): \((\dot{\boldsymbol{J}}+\boldsymbol{\Omega} \wedge \boldsymbol{J}) \cdot \hat{z} = \boldsymbol{G} \cdot \hat{z}\)

03

Show that \(\omega\) is constant

Now, we can rewrite this component along the axis as: \((\dot{\boldsymbol{J}}_{z}+\boldsymbol{\Omega} \wedge \boldsymbol{J}) = G_{z}\) Since \((\boldsymbol{J})\) is perpendicular to the axis, the term \(\boldsymbol{\Omega} \wedge \boldsymbol{J}\) will not have a component along the axis, and we have: \(\dot{J}_{z}= G_{z}\) The angular velocity \(\omega\) is related to \(J_{z}\), and since it is the only externally applied force on the gyroscope along the axis, the angular velocity \(\omega\) will be constant.

04

Determine the vertical component

To find the equation involving the angle \(\varphi\), we need to take components of the given equation in the vertical direction. This can be done by taking the dot product with the unit vector along the Earth's rotation axis, \(\hat{y}\): \((\dot{\boldsymbol{J}}+\boldsymbol{\Omega} \wedge \boldsymbol{J}) \cdot \hat{y} =\boldsymbol{G}\cdot \hat{y}\)

05

Solve the equation for \(\varphi\)

From the vertical component of the given equation, we can obtain the equation for \(\varphi\) as: \(I_{1} \ddot{\varphi}-I_{3} \omega \Omega \sin \lambda \cos \varphi=0\)

06

Find the stable position

We can use the equation for \(\varphi\) to find the stable position by setting the time derivative of \(\varphi\) to zero and solving for \(\varphi\). In this case, we obtain the stable position when the axis is pointing north, i.e., \(\varphi = 0\).

07

Determine the period of small oscillations

To find the period of small oscillations about the northward direction, we can linearize the equation for \(\varphi\) near the stable position \(\varphi=0\). This gives: \(I_{1} \ddot{\varphi}-I_{3} \omega \Omega \sin \lambda = 0\) We can now solve it as a simple harmonic oscillator equation with angular frequency \(\omega_{0}=\sqrt{\frac{I_{3}\omega\Omega\sin\lambda}{I_{1}}}\), and the period will be \(T=2\pi/\omega_{0}\). Substituting the given values of latitude, spinning rate, and gyroscope shape, we can calculate the period for small oscillations.

08

Discuss the sensitivity to horizontal component of Earth's rotation

Since the Coriolis force is responsible for the couple experienced by the gyroscope, the system can be sensitive to the horizontal component of \(\boldsymbol{\Omega}\). From the inertial observer's point of view, the Coriolis force is a fictitious force due to the Earth's rotation, and it influences the motion of the gyroscope. With varying horizontal components of Earth's rotation, the oscillation period and direction of the gyroscope will change, depending on the latitude. Through these steps, we analyzed the motion of the gyroscope and found its relation to the Earth's frame and the Coriolis force. We determined the stable position and period of small oscillations, as well as discussed the sensitivity of the system to the horizontal component of Earth's rotation.

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