Help with a Centripetal Force Question

[NamNodrog]

Homework Statement

On a midway ride called The Round Up, participants climb into a cylindrical apparatus and stand upright with their backs against the wall of the cylinder. The apparatus begins to spin, first in the horizontal plane and then tipping into the vertical plane so that the riders’ bodies are parallel to the ground below. Consider a rider of mass 60.0 kg. The radius of the apparatus is 8.0 m and it spins with a period of 4.0 s.

c. What minimum speed must the person be moving with in order to keep from falling away from the side of the cylinder when rotating in the vertical plane? State where in the circle this would occur and draw the appropriate free-body diagram.

Relevant Equations

Fc=mv/r, v=2pi/r

I am not too sure as to how to approach part c. of this question. In the vertical plane, the centripetal force is provided by the normal force and the force of gravity. However, the solution to this problem includes a description of the forces at the top of the loop, where the normal force is zero, so that g=v^2/r. I don't understand why there is no normal force at the top of the loop.

Thank you for your help.

 

[PhanthomJay]

There is a normal force when rotating above the critical speed. But what would be the normal force on the person when they are about to fall down?

 

[NamNodrog]

There is a normal force when rotating above the critical speed. But what would be the normal force on the person when they are about to fall down?

I thought in this situation there is always an inward normal force. If there is no normal force at the top and the centripetal force is constant, then the centripetal force must be mg (which it is not). Why does the centripetal force change at this point?

Thank you for your response

 

[PeroK]

I thought in this situation there is always an inward normal force. If there is no normal force at the top and the centripetal force is constant, then the centripetal force must be mg (which it is not). Why does the centripetal force change at this point?

Thank you for your response

This is why you need to do the actual calculations. But, note that if there were always a normal force (no matter how slowly the cylinder was spinning) then you could never fall off.

Also, why do you say that the centripetal force cannot be $mg$? Why can't it have that value? As you spin faster, the centripetal force required increases from 0 and has no upper limit. At some rotational speed the force required must just happen to equal $mg$.

 

[NamNodrog]

This is why you need to do the actual calculations. But, note that if there were always a normal force (no matter how slowly the cylinder was spinning) then you could never fall off.

Also, why do you say that the centripetal force cannot be $mg$? Why can't it have that value? As you spin faster, the centripetal force required increases from 0 and has no upper limit. At some rotational speed the force required must just happen to equal $mg$.

I didn't include part a) in the initial question, and it required you to calculate the normal force at the bottom of the circle (which required you to first find the centripetal force) and it did not equal mg.

I just don't understand how in part c) the solution guide begins with Fn=0 at the top of the circle and I don't know why that is the case for this specific question. The question is question 1 here, and you can see the solution by clicking "check work" at the bottom of the page.
http://pages.lockerby.net/coulasb/L... Chapter Reviews/Lesson 1 15 Res/problems.htm

 

[RPinPA]

Anywhere in a circle, you need something to provide force $mv^2/r$, directed toward the center, in order to follow the circular path.

Let's say that number was 10 N. So at the top of the circle, you need 10 N of downward force in order to follow the circular path.

Now let's say gravity $mg$ was providing 9 N. Where's the rest of the force coming from to move you in a circle? From the walls of the container, which therefore need to provide 1 N. That's the normal force in that situation. The normal force is downward, gravity is downward, together they add to 10 N, and you need 10 N to stay in the circle.

Now suppose gravity was providing 9.9N? How much of the 10 N would be provided by the walls of the container?

Now suppose gravity was providing 10 N? What is the normal force now?

And what happens if gravity is providing 11 N?

At the bottom of a vertical circle, the net force to keep you moving in the circle is upward. Again, let's say that you need an upward force of 10 N at the bottom to stay in the circle. And let's say that gravity is pulling down with 10 N. What will ensure that the net force is upward at 10 N? What will provide the difference and how much force is it providing?

That's why the top and the bottom are different. The centripetal force is with gravity at one place and opposite to gravity at the other.

 

[hutchphd]

And what happens if gravity is providing 11 N?

Then you had better speed up the roundup. Happily the required minimum speed is the same for all masses so hopefully the operator is competent and has the speed high enough..

 

[PhanthomJay]

The NET centripetal force is constant. The riders weight mg is constant. So is N constant?? What about at the 90 degree point in the turn?

 

[PeroK]

I didn't include part a) in the initial question, and it required you to calculate the normal force at the bottom of the circle (which required you to first find the centripetal force) and it did not equal mg.

I just don't understand how in part c) the solution guide begins with Fn=0 at the top of the circle and I don't know why that is the case for this specific question. The question is question 1 here, and you can see the solution by clicking "check work" at the bottom of the page.
http://pages.lockerby.net/coulasb/L... Chapter Reviews/Lesson 1 15 Res/problems.htm

The total centripetal force must be the same throughout a rotation, but gravity acts in different "directions": at the bottom gravity is away from the centre; at the sides gravity is tangential to the walls; and at the top gravity is towards the centre. The normal force from the walls must be changing throughout: greatest at the bottom and reducing to a minimum at the top.

If the rotation is too slow, then gravity is too strong at the top and you fall off. The slowest the rotation can possibly be is when gravity provides all the required centripetal force at the top: any slower and you fall off; any faster and you need a normal force as well.

That defines the condition for minimum rotation: $F_c = mg$ and $F_N = 0$ at the top. And, incidentally, $F_N = 2mg$ at the bottom.