Two boys on a merry go round

In summary, the conversation discusses the Coriolis effect and the trajectory of a ball from the perspective of B and someone outside. It is mentioned that the acceleration with respect to B is 0, but this is contradictory as a non-straight path would require an acceleration. The task asks for the acceleration of the ball with respect to B, and it is determined that it only has normal acceleration due to the constant velocity, and it points outside the merry-go-round. However, there is a discrepancy regarding the speed as the Coriolis force does not affect it, but the centrifugal pseudo-force can.
  • #1
Like Tony Stark
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6
Homework Statement
Two boys are sitting diametrically opposite on a merry go round. The platform has a constant angular velocity. Then, the boy ##A## throws a ball to the boy ##B##. This ball has a horizontal velocity relative to the platform and directed to ##B##, and has no horizontal acceleration. What's the acceleration with respect to ##B## along the plane of the platform at the instant after thrown?
Relevant Equations
##\vec a=\vec a_B + \vec{\dot \omega} \times \vec r + \vec \omega \times (\vec \omega \times \vec r) + 2. (\vec \omega \times \vec v_{rel}) + \vec a_{rel}##
This is a tipical Coriolis effect example, and I know what the trajectory drawn by the ball looks like from the perspective of B, and from the perspective of someone outside. But the acceleration with respect to ##B## is 0, isn't it? Because the problem says "the ball has no horizontal acceleration". If not, what would it be? I guess it would be an acceleration pointing outside the platform because that would be the acceleration that moves the ball from the perspective of ##B##
 
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  • #2
Like Tony Stark said:
Coriolis effect example, and I know what the trajectory drawn by the ball looks like from the perspective of ##B##, and from the perspective of someone outside. But the acceleration with respect to ##B## is 0,

Do you see why the bolded phrases are contradictory? If ##B## perceives a path which is not straight in his frame of reference, then by definition the velocity can not be constant.

 
  • #3
RPinPA said:
Do you see why the bolded phrases are contradictory? If ##B## perceives a path which is not straight in his frame of reference, then by definition the velocity can not be constant.

I didn't express myself correctly. Yes, it has an acceleration because it doesn't follow a straight line, so it must have an acceleration. But how would that acceleration be? Because the task says "what's the acceleration of the ball with respect to ##B##?#, which are its characteristics? It only has normal acceleration and not tangential because the velocity is constant, and it points outside the merry go round. Would that be ok?
 
  • #4
Like Tony Stark said:
I didn't express myself correctly. Yes, it has an acceleration because it doesn't follow a straight line, so it must have an acceleration. But how would that acceleration be? Because the task says "what's the acceleration of the ball with respect to ##B##?#, which are its characteristics? It only has normal acceleration and not tangential because the velocity is constant, and it points outside the merry go round. Would that be ok?
I am having difficulty following the logic here.

We have already agreed that the velocity is not constant. So it seems that you mean that the speed is constant because Coriolis is the only force in play. Indeed, Coriolis is always perpendicular to the velocity vector and does not affect speed. But that ignores the centrifugal pseudo-force. Centrifugal force can affect speed.
 

What is the concept of "Two boys on a merry go round"?

The concept of "Two boys on a merry go round" is a classic physics problem that involves two boys of different weights sitting on opposite sides of a merry go round and rotating at a constant speed. This problem is commonly used to explain concepts such as centripetal force, inertia, and angular velocity.

How does the weight of the boys affect the motion of the merry go round?

The weight of the boys affects the motion of the merry go round by creating an imbalance in forces. The heavier boy exerts a greater downward force on his side of the merry go round, causing it to rotate faster. This creates a difference in centripetal force, which pulls the merry go round towards the heavier boy's side.

What is the relationship between the boys' distance from the center and their angular velocity?

The relationship between the boys' distance from the center and their angular velocity is inverse. This means that the farther away the boy is from the center of the merry go round, the slower his angular velocity will be. This is because the boy farther away has a longer distance to travel in the same amount of time as the boy closer to the center.

How does the speed of the merry go round affect the boys' centripetal force?

The speed of the merry go round does not affect the boys' centripetal force. The centripetal force remains constant regardless of the speed of the merry go round. This is because the centripetal force is determined by the boys' weight and their distance from the center, both of which do not change with the speed of the merry go round.

What happens to the boys' motion if they switch positions on the merry go round?

If the boys switch positions on the merry go round, the motion of the merry go round will not change. This is because the centripetal force is determined by the boys' weight and their distance from the center, not their positions on the merry go round. However, the boys' angular velocity will change as the heavier boy will now be closer to the center and the lighter boy will be farther away from the center, resulting in a change in their angular velocity.

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