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- Thread starter Ronnin
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OlderDan

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Your statement in bold is false. You need to rethink everything that follows it.Ronnin said:This is a case where I know the results but am fighting the math.At the top of a ferris wheel your acceleration in Y is 0.

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OlderDan

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The centripetal force is not an additional force in the problem. It is the name we give to the combination of forces that are causing the circular motion. The only forces acting on you are gravity and the normal force, and these are in opposite directions. You know that gravity is stronger, because the acceleration is down toward the center of the circle. If you take down as positive (the direction of the centripital force) you haveRonnin said:

Ftotal = mg - Fn = mv^2/r , where everything represented as a letter is positive

An alternative way of looking at the problem is to treat mv^2/r as a centrifugal force (I don't recommend it) If you do this the direction of mv^2/r is upward, as is the normal force, and gravity is downward. In this case, the net force has to be zero, so you have

mg = Fn + mv^2/r

Either way you do it, you get the same result

Fn = mg - mv^2/r

You were right to question what you wrote earlier as the given answer. Writing the mv^2/r with the + and treating the term as a negative is confusing. There is no natural way to introduce a negative into this quantity.

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Doc Al

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Both weight and centripetal acceleration point downward and are negative.Ronnin said:I still have 2 force vectors working in the downward direction. I've seen the solution but they figure Fn=mv^2/r + mg, but they use a negative for the angular acceleration but a possitive for gravity. That just doesn't make since if they both point toward the center of the wheel.

First identify the forces:

(1) Normal force, which points upward: Fn

(2) Weight, which points downard: -mg

Now add them up:

Total force= Fn -mg

The acceleration is centripetal and points down, so: a = -v^2/r

Now apply Newton's 2nd law (total force = ma):

Fn -mg = -mv^2/r

Thus: Fn = mg -mv^2/r

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Doc Al said:Both weight and centripetal acceleration point downward and are negative.

First identify the forces:

(1) Normal force, which points upward: Fn

(2) Weight, which points downard: -mg

Now add them up:

Total force= Fn -mg

The acceleration is centripetal and points down, so: a = -v^2/r

Now apply Newton's 2nd law (total force = ma):

Fn -mg = -mv^2/r

Thus: Fn = mg -mv^2/r

But why is the centripetal acceleration not accounted for in the total force? It is acting on the mass in the downward direction just as gravity is, right?

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OlderDan

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No. The centripetal force is not an additional force. It is the name we give to the net force acting on an object in circular motion with constant speed. That net force comes from the real forces acting on an object, such as gravity, friction, normal force. When you write the net force in terms of the contributing forces, you know that its magnitude must be mv^2/rRonnin said:But why is the centripetal acceleration not accounted for in the total force? It is acting on the mass in the downward direction just as gravity is, right?

Edit: I added the "with constant speed". An object in circular motion could be accelerating tangentailly as well, but that is not part of this problem. The net force acting on such an object would have both a centrifugal force component and a tangential force component.

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Doc Al

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Thank you guys so much. That was exactly the point I needed clarification on.

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OlderDan

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OOPs That should have saidOlderDan said:The net force acting on such an object would have both a centrifugal force component and a tangential force component.

The net force acting on such an object would have both a

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