# Circular Motion Help

1. Nov 28, 2004

### HawKMX2004

Circular Motion Help PLZ!!!

I have a couple questions, the first:

1.) An 80 kg man sits 9 m from the axis of a ferris wheel moving at .4 radians/sec and sits on a scale. What does the scale read at A.) The Top of the Ride, B.) The bottom of te ride, and C.) 32 degrees from the bottom of the ride.

Here is what I have so far; Fc = Force Centripical; r = radius, v = velocity; m = mass

A.The Top of the Ride
Fc = m(v^2) / r V= 9m(.4 rad) = 3.6 m/s
Fc = 80(3.6^2) / 9
Fc = 115.2 N
F = ma
m = F / a a = v^2 / r = 3.6^2 / 9 = 1.44 m/s/s
m = 115.2 N / 1.44m/s/s
m = 80 kg <--- Answer ( Doesnt seem Right )

B.The Bottom of the Ride
bring down Fc = 115.2 N
Fc = 115.2 N
F = ma
m = F / a a = v^2 / r = 3.6^2 / 9 = 1.44 m/s/s
m = 115.2 N / 1.44m/s/s
m = 80 kg <--- Answer ( Doesnt seem Right Either)

C. 32 degrees above the bottom; 0 = theta

Sin(0) = O/H = O/9 = Sin(30)(9) = O = 4.5 m

Dont know where to go from there.

Second Question

2.) The friction cofficient of static friction between a childs sneakers and a spinning platform that she is standing on is .42 The kinetic coefficient is .35 If the child weighs 28 kg and is standing 1.5 m from the center of the platform, A.) at what velocity will the child start to slide away from the center, B.) At what distance from the center will the child stop sliding

A.) Start Sliding
V = Ac - Ff Ac = v^2/r, Us = .42
V = V^2 /1.5 - Fn(.42)
V^3 = 1.5 - ma(.42)
V = cuberoot(1.5 - 28(9.8)(.42))
V = cuberoot(-113.7)
V = -4.8 m/s <--- Answer ( doesnt seem right )

B.) Distance before stop.

I have no clue, Please Help, All help would be appreciated..i am very very confused with this stuff, please help

2. Nov 29, 2004

### Diane_

With respect to your first question: Let's assume for the sake of argument that the scale reads in Newtons, not kilograms - this being, of course, because the scale is actually measuring force and not mass. If your teacher wants you to express it in mass terms, just divide the answer by 9.80 m/s^2.

Now - Suppose the man weren't moving. In that case, the scale would read the same at the top and the bottom of the circle. 80kg * 9.80 m/s^2 = 784 N.

Now start the ferris wheel moving. Why would the scale reading change? Well, there's good old inertia - since he's moving, he's going to tend to move in a straight line, and the ferris wheel (via the seat belt in his gondola) has to provide a force to keep him going in a circle. This is the centripetal force, which is directed towards the center of the circle. The top and the bottom are special cases, as gravity is directed entirely parallel to or antiparallel to the necessary centripetal force.

At the bottom of the circle, the ferris wheel has to do two things: 1) counteract gravity and 2) provide all of the centripetal force to hold him on. So the scale will read the sum of his weight (the aforementioned 784 N) and the centripetal force, which you can get from mv^2/r.

At the top of the circle, the ferris wheel still needs to counteract his weight, but there's a catch - some of his weight is providing the centripetal force. For all practical purposes, he's trying to fly off of the wheel, but gravity keeps him on it. You can get the reading on the scale by taking his weight and subtracting that part of it that is being "used" for centripetal force. If the wheel were rotating fast enough, the scale would show no force at all, and he would be essentially in free fall, but moving to the side (tangential to the wheel) fast enough that he'd simply coast on over and follow the path of the wheel. If it were turning faster, then the weight alone wouldn't be enough to hold him down, and the seat belt would have to act to hold him in his seat - the scale would show a "negative force", if that makes sense.

Now - assuming the above makes sense, think about what happens as the wheel turns from him on the bottom to him on the top. Clearly, the force provided by the scale will decrease from weight+centripetal force to centripetal force-weight. It will decrease because we're moving away from a place where the weight acts directly away from the center to a place where it acts directly towards the center. What you need to find to answer the third question is the component of the weight that acts towards or away from the center of the wheel. If the component acts towards the center, then it will provide some of the centripetal force, and you'll get the scale reading by subtracting that component from the necessary centripetal force. This will occur on the upper half of the cycle. If the component is acting away from the center (bottom half of the cycle), then you'll add it to the necessary centripetal force.

Does this make sense?

On the second: Picture the child standing on the platform (a merry-go-round?) before it starts moving. There's obviously no tendency for her to move outward. As the platform starts to rotate, there will be an increasing tendency for her to slide. She will not move initially because static friction is enough to keep her moving in the circle - it's enough to provide the necessary centripetal force. At some point, though, static friction will give up the ghost. This will be the place where static friction is just barely not enough to hold her in place. So, to find it, find the speed at which static friction is exactly what's needed to hold her in place. (This should be easy, since we assume the platform is level, so the normal force is just her weight.) The actual answer to the question is "anything above this speed", but I'm sure your teacher will accept that speed as the answer.

For the second one, we're going to have to make an assumption. Consider the case: the platform has reached a speed sufficient to make the child start to slide. As soon as that happens, we're in a new regime. Static friction no longer applies, kinetic friction does. So the force holding her in place drops suddenly, and she starts to move. As she slides outward, the necessary centripetal force drops, and eventually she reaches a place where kinetic friction is enough to keep her in the circle. At that point, she will begin to decelerate, eventually coming to a stop - static friction rears its sticky little head, and she stays there.

The problem is in figuring how far she slides. Since she is still moving outward, the necessary centripetal force will continue to decrease, while kinetic friction will remain constant. It's possible to figure how far out that is, but I don't see any way to do it without integration, and I get the impression that you're not in a calculus-based physics class. My suggestion would be to find the radius for which the kinetic friction on the girl is equal to the centripetal force necessary to hold her in the circle and simply report that - pretty much the same procedure as the first, except that you're dealing with a smaller number and you're solving for r not v. The answer will be a lie, but not too bad a one, I think.

3. Nov 29, 2004

### HawKMX2004

Ok, so for the first one, if i understood you correctly, i take his weight and subtract the force needed for the centripetal force. 80 - 80 = 0..so his weight would be nothing at the top. And at the bottom, i take his weight and add the force needed for the centripetal force. 80 + 80 = 160..so his weight would be 160 kg at the bottom. That still doesnt really make sense, but i think i get what your trying to say.

For the 32 degree from the bottom, i still have no idea how to get that, i dont know what equations to use or anything for that one, do you have any suggestions?

As for the "merry-go-round" i assume i take the velocity at which she'd move without friction (force centripetal) and then subtract Ff (force of friction).

So the equation would look like this V = Fc - Ff ?? I'm not quite sure about that either but i worked it out, and dont know how to solve for Fc because it has V^2 in it could you please help me a little more there?

Your help is greatly appreciated, and i think i might be starting to get it, or really not getting it.. , but thanks for your help