How Does the Coriolis Force Affect a Ball Toss on a Merry-Go-Round?

In summary, the child has to toss the ball at a speed that is faster than the speed of the merry-go-round and at a angle such that the ball reaches the right neighbour.
  • #1
Marvin94
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Homework Statement


Four children are playing toss on a merry-go-round which has a radius of r=2m. The merry-go-round turns counterclockwise and completes one revolution in 2 seconds. The child who has the ball wants to toss it to its right neighbour. It tosses the ball towards the center of the merry-go-roung.

Homework Equations


a) How fast does the child have to throw the ball?
b) How fast does the merry-go-round have to turn to let the child catch its own ball, if it was thrown at the same speed as in a)?
Solve the problem in the (x,y) plane of the carousel, from an outside, resting frame of reference, neglecting gravity and friction.

The Attempt at a Solution


I saw at the beginning of someone's solution that:
[itex]x(t) = x_0 + v_x t + \frac{1}{2} a_x t^2[/itex]
[itex]y(t) = -r + v_y t [/itex]
I don't understand why it should be correct. However I would know if someone could directly explain me a nice approach to this problem and solve it to make it clear to me. Thanks a lot.
 
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  • #2
As I read the problem, we do not need to worry about the vertical dimension. We can work in two dimensions. The proposed equation for x(t) involving ##\frac{1}{2}a_xt^2## is not appropriate. We do not have a constant acceleration situation.

You are told to solve the problem using the non-rotating frame of reference. In that frame the you should be able to compute launch velocity based on the speed with which the thrower tosses the ball. In the non-rotating frame, will the path taken by the thrown ball be straight or curved? Can you figure out where its path will intersect with the rim of the merry-go-round?
 
  • #3
I think the path of the ball from outside will appear only along the y-axis (back and forth).
And we can imagine the angle between the two children to be given.
 
  • #4
Marvin94 said:
I think the path of the ball from outside will appear only along the y-axis (back and forth).
And we can imagine the angle between the two children to be given.
The ball is launched with a velocity relative to the moving child that is purely in the y direction.
 
  • #5
Yes, this is what I meant..
 
  • #6
Marvin94 said:
Yes, this is what I meant..
Given a launch speed and the fact that the launch angle relative to the throwing child is along the y axis, can you compute the speed and direction of the throw in the motionless reference frame?
 
  • #7
We said that the direction of the throw in the motionless reference frame is along the y-axis, right?
The speed depends on where is the other child on the circle. But I don't get the relation between the speed and the other child's position..
 
  • #8
It does simplify the problem greatly if you make the assumption that the direction is along the y-axis of the motionless frame. My reading was that the direction is [initially] along the y-axis of the rotating frame.
 
  • #9
Is there along y-direction some kind of acceleration to be considered?
 
  • #10
From your original post [emphasis mine]
Solve the problem in the (x,y) plane of the carousel, from an outside, resting frame of reference, neglecting gravity and friction
 
  • #11
I was not talking about that quantities... I mean, if from outside point of view the ball moves toward positive y-axis, and then toward negative y-axis, maybe some kind of acceleration which modifies our velocity could be considered. However, if the speed is constant, we will see the ball reaching a max value S along y-axis and then again toward zero. It means the ball travel a distance of 2S, right? So Speed = 2S / time..
 
  • #12
Marvin94 said:
Is there along y-direction some kind of acceleration to be considered?
No.
 
  • #13
Marvin94 said:
I was not talking about that quantities... I mean, if from outside point of view the ball moves toward positive y-axis, and then toward negative y-axis,
Where do you get the idea that the ball moves back and forth?

However, if the speed is constant, we will see the ball reaching a max value S along y-axis and then again toward zero.
What is S supposed to be? No S is defined in the problem. No S is defined anywhere in anything you have written.
 
  • #14
If the ball go toward the center of the circle, then it has to come back to reach the other child. That's why I thought the ball moves toward positive y-axis first, and then toward negative y-axis.
S is not defined.. I just set it as the half "right distance" to walk in order to reach the other child.
How would you solve the task?
 
  • #15
Marvin94 said:

Homework Statement


Four children are playing toss on a merry-go-round which has a radius of r=2m. The merry-go-round turns counterclockwise and completes one revolution in 2 seconds. The child who has the ball wants to toss it to its right neighbour. It tosses the ball towards the center of the merry-go-round.

Homework Equations


a) How fast does the child have to throw the ball?
b) How fast does the merry-go-round have to turn to let the child catch its own ball, if it was thrown at the same speed as in a)?
Solve the problem in the (x,y) plane of the carousel, from an outside, resting frame of reference, neglecting gravity and friction.

The Attempt at a Solution


I saw at the beginning of someone's solution that:
[itex]x(t) = x_0 + v_x t + \frac{1}{2} a_x t^2[/itex]
[itex]y(t) = -r + v_y t [/itex]
I don't understand why it should be correct. However I would know if someone could directly explain me a nice approach to this problem and solve it to make it clear to me. Thanks a lot.
As briggs pointed out, that "someone's" solution is wrong. -- no need to involve acceleration . This assumes that we don't need to consider the vertical direction. This allows the problem to be worked out in two dimensions. If the vertical direction is to be included, that can be worked out after the two dimensional solution is completed.

Some suggestions:
Pick a location in the resting frame to place the center of the merry-go-round. -- maybe at the origin?

Pick a location in the resting frame at which the child tosses the ball. Since the discussion so far has suggested that the child attempts a toss along the y-axis, I suggest a location of (0, -2) in the resting frame. I also suggest that when the child is at this location, the rotating reference frame (fixed to the merry-go-round) is aligned with the resting frame.

Now, for some questions:

What is the velocity of the child (in the resting frame) when at location (0, -2) ?
 
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  • #16
According to the given information the velocity of the child (which is tangential to the circle, due to its motion) should be [itex]2 \pi[/itex]
 
  • #17
Marvin94 said:
According to the given information the velocity of the child (which is tangential to the circle, due to its motion) should be [itex]2 \pi[/itex]
(in units of m/s.) Right?

If the child attempts to toss the ball to the center, he/she will give the ball a velocity that only has a component in the y-direction (in the rotating frame). But, in the resting frame, what is the x-component of that velocity?
 
  • #18
Yes m/s.
This point is not very clear to me... why should the velocity from the resting frame have a x-component?
 
  • #19
Marvin94 said:
Yes m/s.
This point is not very clear to me... why should the velocity from the resting frame have a x-component?
In the rotating frame, neither the child, nor the ball have an x-component of velocity at the moment it's tossed.

But, in the resting frame the child does an x-component of velocity at that moment. (You just told me what it is.) So the ball must have that same component with respect to the resting frame. Right?
 

FAQ: How Does the Coriolis Force Affect a Ball Toss on a Merry-Go-Round?

1. What is the Coriolis force on a playground?

The Coriolis force on a playground is a fictitious force that appears to act on objects moving on a rotating surface, such as the Earth. It is caused by the rotation of the Earth and can affect the direction of moving objects.

2. How does the Coriolis force affect objects on a playground?

The Coriolis force can affect the direction of moving objects on a playground, causing them to veer to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. This is because the Earth's rotation creates a deflection of moving objects towards the right in the Northern Hemisphere and towards the left in the Southern Hemisphere.

3. What causes the Coriolis force on a playground?

The Coriolis force on a playground is caused by the rotation of the Earth. As the Earth rotates, objects on its surface are also rotating, and this rotation creates a deflection force known as the Coriolis force.

4. How does the rotation of the Earth affect the Coriolis force on a playground?

The rotation of the Earth is what creates the Coriolis force on a playground. The magnitude of the Coriolis force is directly proportional to the speed of rotation of the Earth, which is why it is more noticeable at higher latitudes closer to the poles.

5. What are some real-life examples of the Coriolis force on a playground?

The Coriolis force on a playground can be observed in many real-life situations, such as the movement of air masses and ocean currents, the trajectory of flying objects like planes and missiles, and the direction of water draining from a sink or bathtub. It can also affect the trajectory of objects on a merry-go-round or other rotating playground equipment.

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