Carnival ride problem (horizontal circular motion)


by xshadow93x
Tags: carnival, circular motion, horizontal
xshadow93x
xshadow93x is offline
#1
Apr3-09, 09:05 PM
P: 2
1. The problem statement, all variables and given/known data

A carnival ride spins the riders in a horizontal circle. During the ride the floor falls away from the riders. The average mass of a rider is 110.6 kg and the ride has a radius of 18.2 m. The coefficient of static friction between the riders and the wall is 0.2400. What is the minimum tangential speed the ride must maintain in order to prevent the riders from falling?

m = 110.6 kg
r = 18.2 m
μs = 0.2400
v = ???

2. Relevant equations

Fc = (m*v2)/r
Fc = m*g
Fs = μs*FN

3. The attempt at a solution

I tried doing this:

1.) Finding FN by multiplying the mass by 9.8.
2.) Plugging this number into Fs = μs*FN
3.) Subtracting Fs from FN to get Fc after friction had been factored in. (I am unsure of these first few steps.)
4.) Plugging in all currently known variables into the equation Fc = (m*v2)/r to solve for velocity coming up with 11.643 m/s. However using this method I have been getting these type of questions wrong on my online practice quizzes.
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turin
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#2
Apr3-09, 09:16 PM
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Draw a free-body-diagram. You should be able to determine FN. BTW, you have two different Fc's; is there a reason for this?
xshadow93x
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#3
Apr3-09, 09:30 PM
P: 2
Quote Quote by turin View Post
Draw a free-body-diagram. You should be able to determine FN. BTW, you have two different Fc's; is there a reason for this?
Would FN be the force of the wall pushing in? And what would I then do with that force in relation to the equations? I have a feeling I have multiple mistakes in my work.

turin
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Apr3-09, 11:44 PM
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Carnival ride problem (horizontal circular motion)


How did you determine your first "Fc"?

Again, I think you should draw a fbd.
JazzFusion
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#5
Jun24-09, 02:39 PM
P: 96
Quote Quote by xshadow93x View Post
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3. The attempt at a solution

I tried doing this:

1.) Finding FN by multiplying the mass by 9.8.
This is a very common mistake and your first problem. The normal force is not necessarily equal to mg in all cases. The normal force FN is the 'equal and opposite' force out that balances a force in to any surface. It is a vector (has magnitude and direction), so that when you add the normal force vector and the vector of the force into a surface, they balance (the sum of the two vectors equal zero). Otherwise, if one or the other force is larger, some object will jump off a surface, or be pushed through it.

The normal force may be mg, but it doesn't have to be. The important thing is that it is exactly what it needs to be (no more and no less) to balance some other force. Take the example of a book sitting on a flat surface (like a table). Gravity pulls the book down mg, so the normal force on the book must be mg up. Now push down on the book with an additional force. The total force down on the book is mg + Fpush, so the normal force acting on the bottom of the book must now be (mg + Fpush) up (otherwise, if the normal force doesn't balance this additional force, I will succeed in pushing the book straight through the table!).

As a further example, take that same book and push it hard, straight into a vertical wall. Now the normal force = Fpush out of the wall (to balance exactly my force Fpush in to the wall). Notice that the normal force now doesn't have anything to do with gravity. Gravity is also acting on the book, however, an amount mg down. If the book is not moving, then this force must be balanced by some other force, equal to mg up. Can you guess what would provide the necessary balancing force against gravity, to keep the book from moving in this case? Think about it for a moment, and then roll over the following text to see if you guessed correctly.

Spoiler
[The force that balances gravity in this case is static friction. The direction of the force is up, against gravity.]


The case of the book pushed against the wall is almost exactly the same as the problem you are considering. The same forces apply, except you don't have a force pushing into the wall (the forces are not balanced, which is why there is a net acceleration). It is the normal force that acts as a 'centripetal force' to hold the rider on a circular path. VERY IMPORTANT CONCEPT: Please remember that 'centripetal force' is a type of force, and not the name of some specific force. 'Centripetal' means 'in towards the center'. So, any force that points 'in towards the center' to keep something moving in a circular path is a centripetal force. Centripetal force can be provided by gravity (it keeps planets moving in orbits), tension (it keeps a rock swung in a sling moving in a circle), or the normal force (as in this case). Centripetal force = m (centripetal acceleration) = mv^2/r.

Try using that expression for the normal force, and see if you can tackle the problem.


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