Verify Coriolis Acceleration Derivation on Merry Go Round

In summary, the conversation is discussing a derivation for the coriolis acceleration on a "merry go round" from the rotating point of view. There is some confusion about the final equation, but it is eventually determined that a 2 was missing from the first term and the unit vector was incorrect in the second term. It is suggested to also consider solving for the trajectory, rather than just the acceleration.
  • #1
somebodyelse5
37
0
I have a derivation for the coriolis acceleration on a "merry go round" that my instructor gave in class, i was wondering if someone could tell me if this is correct or offer the correct final equation.
Additional information: this is from the rotating point of view on the merry go round.

[[tex]\omega [/tex] V[tex]_{}o[/tex]cos([tex]\theta[/tex]-[tex]\omega[/tex]t)+[tex]\omega[/tex]^2V[tex]_{}o[/tex]tsin([tex]\theta-\omega[/tex]t)]j-2[[tex]\omega[/tex]V[tex]_{}o[/tex]sin([tex]\theta[/tex]-[tex]\omega[/tex]t)-[tex]\omega[/tex]^2V[tex]_{}o[/tex]cos([tex]\theta[/tex]-[tex]\omega[/tex]t)]j

Note: this is not HW, just something i need to study for my exam and I am concerned it is incorrect.
V[tex]_{}o[/tex] is Initial velocity

Please forgive the poor formatting of that equation, i did my best to make it clear. For some reason, some of the omegas are off set up, and shouldn't be, could figure out why.
 
Physics news on Phys.org
  • #2
This is what you wrote I think,

[tex]
[\omega v_0\cos(\theta-\omega t)+\omega^2v_0t\sin(\theta-\omega t)]\mathbf{j}-2[\omega v_0\sin(\theta-\omega t)-\omega^2 v_0\cos(\theta-\omega t)]\mathbf{j}
[/tex]

You can just use the tex environment from now on instead of switching between them. Well the way it's written it can't be right since the last term (the one with cosine) doesn't have the units of acceleration. I imagine there should be a [itex]t[/itex] multiplied at the end there.

Did you mis-transcribe it?
 
  • #3
The coriolis force produces acceleration:

[tex]\bold{a_c}=-2(\bold{\omega} \times \bold{v})[/tex]

So, without knowing how exactly you defined your variables, and the exact set up of your problem, I'm not entirely sure I can help you all that much.
 
  • #4
appreciate the responses guys. In my frantic searches and calls to fellow students it turns out i simply "forgot" a 2 in front of the first term, and mixed up the unit vector for the second term.

If you guys are at all interested, I can post the correct equation tomorrow? got to keep studying now!
 
  • #5
You are missing a time variable in the second centrifugal acceleration term. You need a v*t to equal radius for someone walking radially. The 2 should only be part of the Coriolis acceleration and not the centrifugal acceleration. Your second unit vector j should be an i. Other than that you're fine. Also I'm assuming that a positive v means the person is walking radially outward, and a negative v means they are walking radially inward.

Have you tried solving for the trajectory and not just the acceleration?
 

1. What is Coriolis acceleration and how does it affect objects on a merry-go-round?

Coriolis acceleration is the apparent acceleration experienced by an object on a rotating frame of reference such as a merry-go-round. It is caused by the combination of the object's linear velocity and the rotation of the frame. This acceleration causes objects to move in a curved path, rather than a straight line, on a rotating surface.

2. What is the derivation of Coriolis acceleration on a merry-go-round?

The derivation of Coriolis acceleration on a merry-go-round involves using the equations of motion for circular motion and the relative velocity between the rotating frame and the object. This results in a formula for the Coriolis acceleration, which takes into account the angular velocity of the merry-go-round and the linear velocity of the object.

3. How is the Coriolis acceleration different from the Coriolis force?

The Coriolis acceleration and the Coriolis force are closely related but distinct concepts. The Coriolis force is a fictitious force that appears to act on objects in a rotating frame, while the Coriolis acceleration is the actual acceleration experienced by the object due to the rotation of the frame.

4. How does the Coriolis acceleration affect the trajectory of objects on a merry-go-round?

The Coriolis acceleration causes objects to move in a curved path on a rotating surface, as it is a result of the combination of linear and rotational motion. This means that objects will appear to deviate from a straight path and may experience a change in direction as they move around the merry-go-round.

5. What are some real-life examples of the Coriolis acceleration in action?

The Coriolis acceleration is an important concept in meteorology, as it affects the movement of air masses and the formation of weather systems. It is also relevant in sports such as baseball and football, where the rotation of the Earth can impact the trajectory of the ball. Additionally, it plays a role in the motion of ocean currents and in the design of amusement park rides such as the Gravitron.

Similar threads

  • Classical Physics
Replies
9
Views
1K
Replies
14
Views
1K
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Classical Physics
Replies
6
Views
1K
Replies
8
Views
781
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top