Vertical circle in a pendulum ride -- tension force acting on the gondola

[Nikitta]

Homework Statement

In a vertical circle more specifically in a pendulum ride, there is a tension force acting on the gondola and a reaction force by the gondola acting on the passengers. How do you know the tension force and reaction force are equal and how do you find the reaction/tension force at the bottom of the circle?

Relevant Equations

T = mv^2/r + mg
Ep=mgh
V (critical) = SQRT(gr) (velocity at top of circle)
Ek (lin) = 1/2mv^2
Ek (rot) = 1/2Iw^2
Ek (total) = 1/2mv^2 + 1/2Iw^2

At the bottom of the circle, the tension force is greater than the weight force as there must be a net force acting towards the centre to provide the centripetal force causing the centripetal acceleration and thus the circular motion. In the equation above (T = mv^2/r + mg) I only have the mass and radius. I tried to find the velocity of the object at the bottom of the circle by using conservation of energy (Ep lost = Ek gained). I tried to find the total energy of the system, gravitational potential, linear and rotational kinetic energy at the top, however, I could not find the rotational kinetic energy (don't have I). Does the rotational kinetic energy need to be taken into account (In other problems about vertical circle the rotational kinetic is usually not involved, is there any rotational kinetic energy?)

 

[haruspex]

there is a tension force acting on the gondola and a reaction force by the gondola acting on the passengers. How do you know the tension force and reaction force are equal

I wouid think the gondola has mass, and it is undergoing acceleration as well as being subject to gravity , so why would these two forces be equal? Is this the exact wording?

Does the rotational kinetic energy need to be taken into account

Technically, yes, but if you not given any data for that assume you are meant to ignore it.

 

[Nikitta]

I wouid think the gondola has mass, and it is undergoing acceleration as well as being subject to gravity , so why would these two forces be equal? Is this the exact wording?

Technically, yes, but if you not given any data for that assume you are meant to ignore it.

in this video she says that the tension force on the cup (kind of like the gondola) is equal to the reaction force that the cup exerts on the water (kind of like the passengers)

 

[Nikitta]

in this video she says that the tension force on the cup (kind of like the gondola) is equal to the reaction force that the cup exerts on the water (kind of like the passengers)

Also how would you estimate the rotational inertia of an object, since for each object there is a different equation and for some there is no equation (could I just use 2 equations one for a rod and the other for a hoop and add them together?). I have the mass and length of the object.

 

[haruspex]

I have the mass and length of the object.

Please supply all the information you have in your initial post.

in this video she says that the tension force on the cup (kind of like the gondola) is equal to the reaction force that the cup exerts on the water

Only if the mass of the cup is ignored. It is rather more dubious to ignore the mass of a gondola, which may well exceed the mass of the occupants.

 

[Nikitta]

Only if the mass of the cup is ignored. It is rather more dubious to ignore the mass of a gondola, which may well exceed the mass of the occupants.

I think I understand now. If I am calculating the tension force (T=mv^2/r + mg) at the bottom I would use the mass of the gondola and passengers combined, but if I am calculating the reaction force (Fn=mv^2/r + mg) I would only use the mass of the passenger. Is this correct? Thanks

 

[Nikitta]

Please supply all the information you have in your initial post.

Sorry I can't edit the initial post. Mass (gondola + tube) = 12,000 kg, length (tube) = 15 m and the diameter of the gondola = 6 m. I don't think I have enough information to estimate the rotational inertia, I'm assuming that I would need to have the mass of the gondola and tube separate, but thanks for your help.

 

[haruspex]

I think I understand now. If I am calculating the tension force (T=mv^2/r + mg) at the bottom I would use the mass of the gondola and passengers combined, but if I am calculating the reaction force (Fn=mv^2/r + mg) I would only use the mass of the passenger. Is this correct? Thanks

Yes.

Sorry I can't edit the initial post.

I meant, in future.