Gyroscope (Coriolis Torque and Force)

[andyonassis]

Homework Statement

Using integration, show that the total coriolis torque acting on the gyroscope when the wheel has a mass M is:
$$ \vec{N} = -\frac{1}{2}Mr^2\omega\dot{\theta} \sin \alpha \hat{y} $$
This torque is the basis of the gyrocompass.
The Coriolis force can produce a torque on a spinning object.
The $\hat{x},\hat{y}$ and $\hat{z}$ components of $d\vec{N}$ about the origin due to the coriolis force in the $xyz$ coordinate system which acts on a point mass $m$ on the gyroscope's rim is given by:
$$ d\vec{N} = -2r^{2}\omega\dot{\theta}[(\sin(\alpha)cos(\theta)(\sin\theta)\hat{x}-\cos^{2}(\theta) \sin (\alpha) \hat{y})] \times [dm]$$
$\alpha$ is the angle that the gyro makes with the earth and $\theta$ is the angle of the gyroscope that rotates around its axis.

Relevant Equations

Equations for coriolis force and torque.

I know that to prove the total coriolis torque from the coriolis torque on a point mass is to express dm as a function of $d\theta$ and integrate from 0 to $2\pi$ and then the x component disappears due to orthogonality of sine and cosine. But i am stuck at other parts.

 

[NateD]

To prove the total Coriolis torque from the Coriolis torque on a point mass, you need to first start by expressing the torque on a point mass as a function of the angular velocity. The torque on a point mass is given by:$$ \tau_{Coriolis} = m \cdot \omega \times (\omega \times r) $$Where m is the mass of the point mass, ω is the angular velocity, and r is the vector from the origin to the point mass.Next, you need to express the mass of the point mass, m, as a function of the angular displacement, θ. This can be done using the equation for the area of a circle, A = πr2, where r is the radius. Thus, the mass of the point mass is given by:$$ m = \frac{A}{2\pi r} = \frac{r}{2\pi} d\theta $$Now that the mass is expressed as a function of the angular displacement, we can substitute this expression into the expression for the Coriolis torque. This gives us:$$ \tau_{Coriolis} = \frac{r}{2\pi} d\theta \cdot \omega \times (\omega \times r) $$Finally, to find the total Coriolis torque, we need to integrate this expression from 0 to 2π. This gives us:$$ \tau_{total} = \int_0^{2\pi} \frac{r}{2\pi} d\theta \cdot \omega \times (\omega \times r) $$Since the x component of the torque disappears due to orthogonality of sine and cosine, the final expression for the total Coriolis torque is:$$ \tau_{total} = \int_0^{2\pi} \frac{r}{2\pi} d\theta \cdot \omega_y \cdot (\omega_z \cdot r_x - \omega_x \cdot r_z) $$

 

[chrisk]

This is a Symon problem. I found the given expression for the torque is incorrect. Attached is the solution and an explanation of why the solution does not agree with textbook. I know the policy is not to provide solutions but in this case it's necessary to show the error in the stated problem.

 

[berkeman]

I know the policy is not to provide solutions but in this case it's necessary to show the error in the stated problem.

In this case it's a 4 year old thread, so posting the solution is fine. Thanks for updating the thread with the correct solution.