Rotating Frames of Reference question.

In summary: This means that the tension in the string is the same as the force that is trying to pull it towards the center of the circle (which is just centripetal acceleration), and that is why the string makes a perfect circle.
  • #1
John H
36
0

Homework Statement


You are standing on a slowly rotating merry-go-round, turning counterclockwise as viewed from above. You are holding a string from which is suspended a rubber stopper of mass 45g. You are 2.9m from the center of the merry-go-round. You take 4.1s to complete one revolution.

A)Draw FBD of yourself in reference frame of the merry-go-round and earth(assuming to have no rotation).

B)Draw FBD of stopper in Earth's reference frame of a person looking eastward behind you.

C) Draw FBD of the stopper in your reference frame.

D) What angle does the string make with the vertical?

E) What is the magnitude of tension in the string.

Homework Equations



a_c= v^2/r

F_c=(mv^2)/r

Centripetal acceleration is also

a_c= ((4π^2 r)/T^2)

Centripetal force is also

F_c= (M)((4π^2 r)/T^2)

F= ma (Newton's second law)

The Attempt at a Solution



Here is what I attempted so far. Image is pretty big Please go to this url.

http://img801.imageshack.us/f/scan0008r.jpg/"

Thank you in advance.
 
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  • #2
You don't need to use centrifugal force, and certainly not coriolis force, which only applies to something with velocity. Just use the equations for centripetal acceleration to find out how much horizontal force there and compare it with gravity's vertical force to get your angle and tension.
 
  • #3
I got the correct answer using what you told me, but I fail to understand the logic. Help would be appreciated.
 
  • #4
All the points on the outer edge of the circle have a centripetal force that is pointing towards the center of the circle. So the string you are holding has some force pushing it towards the center (you don't go towards the center because friction keeps you on the circle), and gravity has a force pulling the string down (you don't go down because the circle pushes back up on you).
 
  • #5


I would like to provide a response to the content provided:

Firstly, the concept of rotating frames of reference is important in understanding the motion of objects in a non-inertial reference frame. In this scenario, we have a person standing on a rotating merry-go-round, which is a non-inertial reference frame, and a rubber stopper which is suspended by a string. The stopper is also rotating along with the merry-go-round, but it is important to note that its motion is independent of the person standing on the merry-go-round.

Now, let's address the questions asked:

A) Drawing the free body diagram of the person in the reference frame of the merry-go-round, we can see that there are two forces acting on the person: the normal force from the ground and the centrifugal force due to the rotation of the merry-go-round. The normal force is directed towards the center of the merry-go-round, while the centrifugal force is directed outwards, away from the center.

B) In the Earth's reference frame, a person standing eastward behind the person on the merry-go-round would see the stopper moving in a circular motion with a constant speed. The free body diagram of the stopper would show two forces acting on it: the tension force from the string and the force of gravity pulling it down towards the ground.

C) In the reference frame of the person on the merry-go-round, the stopper is at rest. However, since the person is rotating, there is a centrifugal force acting on the stopper, directed outwards, away from the center of rotation.

D) The string makes an angle with the vertical because the stopper is being pulled towards the center of the merry-go-round due to the tension force. This angle can be calculated using trigonometry, and it will depend on the speed of rotation, the radius of the merry-go-round and the length of the string.

E) The magnitude of tension in the string can be calculated using Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. In this case, the acceleration is the centripetal acceleration, which can be calculated using the formula a_c = v^2/r. Therefore, the magnitude of tension in the string can be calculated using the formula F_tension = m(v^2)/r, where m is the mass of the stopper, v is
 

Related to Rotating Frames of Reference question.

1. What is a rotating frame of reference?

A rotating frame of reference is a coordinate system that is rotating with respect to an inertial frame of reference. In this frame, objects appear to experience fictitious forces such as centripetal and Coriolis forces.

2. Why do we use rotating frames of reference?

Rotating frames of reference are used to simplify the analysis of rotating systems, as they eliminate the need for considering fictitious forces. They are also useful for studying celestial bodies and other rotating objects.

3. How do we define the axes in a rotating frame of reference?

The axes in a rotating frame of reference are defined based on the rotation of the frame. The z-axis is always aligned with the axis of rotation, and the x and y axes are perpendicular to the z-axis and each other.

4. What is the difference between a rotating frame of reference and an inertial frame of reference?

An inertial frame of reference is a non-accelerating frame of reference, while a rotating frame of reference is an accelerating frame. In an inertial frame, Newton's laws of motion hold true, but in a rotating frame, fictitious forces must be taken into account.

5. Can we convert between a rotating frame of reference and an inertial frame of reference?

Yes, we can convert between frames of reference using transformation equations. These equations take into account the angular velocity of the rotating frame and the position of the object in the frame to determine its position in the inertial frame and vice versa.

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