Circular Motion involving a gravitron

[Jimmy87]

Homework Statement

I have attached questions about gravitron. I'm struggling with question 2 about calculating the normal force of the gravitron.

Homework Equations

mv^2/r

The Attempt at a Solution

I know the answer so I can figure it out as I know it involves setting the Ncos(62) = mg (N = normal force) but I don't get how you know the angle involved is 62 as it only gives you the angle the track makes with the floor?

Also, please could someone explain why the track is tilted as appose to being vertically upright and what this has to do with enabling the track to slide up when its rotating?

 

[Doc Al]

I know the answer so I can figure it out as I know it involves setting the Ncos(62) = mg (N = normal force) but I don't get how you know the angle involved is 62 as it only gives you the angle the track makes with the floor?

The couch is attached to the track, so it makes the same angle. (Is that your question?) Have you drawn yourself a free body diagram of the person?

Also, please could someone explain why the track is tilted as appose to being vertically upright and what this has to do with enabling the track to slide up when its rotating?

The normal force must have a vertical component to lift the person/couch. If it were upright, the weight could not be supported or lifted.

 

[Jimmy87]

The couch is attached to the track, so it makes the same angle. (Is that your question?) Have you drawn yourself a free body diagram of the person?


The normal force must have a vertical component to lift the person/couch. If it were upright, the weight could not be supported or lifted.

Thanks Doc Al. I have drawn a free body diagram and attached it. I have resolved the normal force into components and labeled what I think each of them do. Regarding the angle, I know from the answer that one of the angles in that vector triangle I have drawn must be 62 degrees but how do you know this? How can you show that the angle between the track and the ground is the same as the angle in that vector triangle?

 

[Doc Al]

Thanks Doc Al. I have drawn a free body diagram and attached it. I have resolved the normal force into components and labeled what I think each of them do.

Good.

Regarding the angle, I know from the answer that one of the angles in that vector triangle I have drawn must be 62 degrees but how do you know this? How can you show that the angle between the track and the ground is the same as the angle in that vector triangle?

The key is to recognize that the angle that the normal makes with the vertical is the same angle that the track makes with the horizontal. That is true because the track and the normal are always perpendicular. One way of seeing that would be to use a bit of triangle geometry or trig. If you label all the angles, I could walk you through it.

Even easier might be this. Imagine that the track was on a hinge that could move through any angle θ, instead of being fixed at 62 degrees. Now lay the track flat so that θ = 0. What angle does the normal make with the vertical? 0, of course, since it points straight up. Now increase the angle to 1 degree. Can you see that the normal must move 1 degree to the vertical? No matter what angle the track makes with the horizontal, the normal will make that same angle with the vertical.

Make sense?

 

[Jimmy87]

Good.The key is to recognize that the angle that the normal makes with the vertical is the same angle that the track makes with the horizontal. That is true because the track and the normal are always perpendicular. One way of seeing that would be to use a bit of triangle geometry or trig. If you label all the angles, I could walk you through it.

Even easier might be this. Imagine that the track was on a hinge that could move through any angle θ, instead of being fixed at 62 degrees. Now lay the track flat so that θ = 0. What angle does the normal make with the vertical? 0, of course, since it points straight up. Now increase the angle to 1 degree. Can you see that the normal must move 1 degree to the vertical? No matter what angle the track makes with the horizontal, the normal will make that same angle with the vertical.

Make sense?

Thanks again Doc Al! Kind of making sense. I have drawn another FBD (attached) and I think I have figured out the angle using a bit of geometry, please could you check to see if it's correct? I have extended the weight force in purple. If the angle between the track and the ground is 62 then the other angle (a) must be 28. I think I am right in saying that because there are two straight lines crossing that a and b are alternate interior angles? Angle c must be 90 minus 28 which is 62 degrees. Is that how you prove it or is that a very long way of doing it?

Also, am I right in saying that the vertical component of the normal force will be greater than the weight force when the couch slides up?

 

[Doc Al]

Thanks again Doc Al! Kind of making sense. I have drawn another FBD (attached) and I think I have figured out the angle using a bit of geometry, please could you check to see if it's correct? I have extended the weight force in purple. If the angle between the track and the ground is 62 then the other angle (a) must be 28. I think I am right in saying that because there are two straight lines crossing that a and b are alternate interior angles? Angle c must be 90 minus 28 which is 62 degrees. Is that how you prove it or is that a very long way of doing it?

That's pretty much it. Here's my version. Call the angle that track makes with the horizontal θ, instead of a specific number. Angles b and c must add to 90, since they make a right angle. Same for angles a and θ. But you know that a = b, so therefore c = θ.

Symbolically:
b + c = 90
a + θ = 90
thus b + c = a + θ
but b = a, so:
c = θ

Also, am I right in saying that the vertical component of the normal force will be greater than the weight force when the couch slides up?

Yes.

 

[Jimmy87]

That's pretty much it. Here's my version. Call the angle that track makes with the horizontal θ, instead of a specific number. Angles b and c must add to 90, since they make a right angle. Same for angles a and θ. But you know that a = b, so therefore c = θ.

Symbolically:
b + c = 90
a + θ = 90
thus b + c = a + θ
but b = a, so:
c = θ


Yes.

Thank you very much!