Conceptual problem involving centripetal accel

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    Centripetal Conceptual
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Discussion Overview

The discussion revolves around a conceptual problem involving centripetal acceleration in the context of an amusement park ride. Participants explore the forces acting on a rider in a rotating cylinder and seek to derive the minimum tangential velocity required for the rider to remain adhered to the wall when the floor drops out. The scope includes theoretical reasoning and mathematical derivation related to forces and friction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem and their initial incorrect derivation, questioning the role of the coefficient of static friction in the final equation.
  • Another participant clarifies the forces acting on the rider, emphasizing the need to consider both vertical and radial forces, and derives the correct relationship between velocity, radius, gravitational acceleration, and friction.
  • A third participant reiterates the force balance, distinguishing between equilibrium in the vertical direction and the necessity of centripetal force in the horizontal direction, while also addressing common misconceptions about centripetal force.
  • The final participant expresses gratitude for the clarification, noting confusion regarding the application of friction in the horizontal versus vertical contexts.

Areas of Agreement / Disagreement

Participants generally agree on the need to analyze both vertical and horizontal forces, but there is no explicit consensus on the initial misunderstanding regarding the role of friction in the problem. The discussion remains somewhat unresolved as participants clarify different aspects of the problem without reaching a unified conclusion.

Contextual Notes

Participants highlight the importance of understanding the distinction between forces in equilibrium and those providing centripetal acceleration. There are indications of confusion regarding the application of friction in different directions, which may affect the derivation of the correct formula.

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amusement park ride...basically a large cylinder rotating about a vertical axis. when it is spinning fast enough the floor is dropped out and the riders stick to the wall well above the floor. find the minimum Vtangential needed. express answer in terms of R of circle, coefficient of static friction (mew)s and gravitational acell (g)... seems easy right, and it probably is, but i don't know why my answer is not right...btw the correct answer is

V= (gR/mews)^.5

here's what i got so far...

x direction forces) V^2/ (r) * m - static coeff *n = 0

y direction forces) static coeff *n - mg = 0


so mg = sf *n so v^2/ R = g so



V = (g/R)^.5 where does the sfriction coeff come from?!?

i worked backwards and saw that is must have been multiplied by
v^2/R...now why would you do that... thanks in advanced
 
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Well this is actually a 2-D problem, the FBD for the rider should be something as follows.

We have the frictional force upwards, a weight force downwards, and a normal force radially inwards.

Now use sumF = ma

In the vertical direction we have
Friction - Weight = 0
u*N = mg

In the radially inward direction we have

Normal = Centripetal acceleration
N = mv^2/r

thus

u*m*v^2/r = m*g

u*v^2/r = g

v = sqrt(r*g/u)
 
vertically, mg=(mew)*N i.e. equilibrium

horizontally, N= mv^2/r.

i.e. not equilibrium, but rather the normal force provides the centripetal force.

substitute N into above, get V=(g*r/(mew))^0.5.

typically, when someone learns centripetal force for the 1st time, they get confused with the whole forces producing the centripetal force.

centripetal force is the net force acting on the object, it is the sum of all external forces. If this net force is not directed perpendicular to the velocity of the object, then the path is not going to be circular.

the object is not in equilibrium on horizontal plane, but is accelerated.

you need to sit back and think deeply about this.
 
thanks a lot guys,,,, what through me off was i thought there was friction on the horizontal, but it seems it only exists on the vertical forces, i really appreciate the help, this site is amazing... thanks for your time
 

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