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Centripetal force carnival ride question

  1. Dec 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 8m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/ min when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?


    2. Relevant equations
    mv^2 / r
    mg

    3. The attempt at a solution
    I tried to convert 45 rev / min into 3/4 rev / sec. From there, I found that it covered 6 pi m / s using the fact that circumference is 2pi r.
    Then I drew a free body diagram with centripetal force, weight, and normal. From there I got lost regarding the relationship between normal and centripetal force.
     
  2. jcsd
  3. Dec 23, 2011 #2

    BruceW

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    you've got the correct speed of 6 pi m/s so you can use this to calculate the centripetal force. And you are right that the next thing to do is to think of the relationship between normal and centripetal force. You need to think of the definition of the centripetal force, and then you'll realise the relationship.
     
  4. Dec 23, 2011 #3
    well, I understand that the centripetal force is responsible for pulling a particle towards the center and keep it in circular motion. I saw that it's equal the normal force but how exactly is that relationship justified?
     
  5. Dec 24, 2011 #4

    BruceW

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    You are correct the centripetal force is equal to the normal force. If you think about the real forces, there is gravity, friction and the normal force. Out of these, which are directed towards the centre of the circle? This will give you the answer to why only the normal force contributes to the centripetal force.
     
  6. Dec 24, 2011 #5
    Centripetal force in NOT a new force that comes into action when a body is rotating with respect to a frame ... it is like a classification of force that is the reason for body's circular motion ... and force could act as centripetal ... Coulomb force in case of atoms, magnetic in particle accelerators, Tension (electric by origin) in case of spinning something tied to a string, normal (electric in nature again) in your case, gravitational in case of planets ... and many more examples ...

    you must work by writing the centripetal force (ie [itex]\frac{mv^2}{R}[/itex]) and equating it to (forces that are helping the rotation of body) - (forces that are against the rotation of body)
     
  7. Dec 24, 2011 #6

    vela

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    To avoid this common misconception, one thing you might want to do is to rid yourself of the notion of a centripetal force. Objects following a curved path experience a centripetal acceleration ##a=v^2/r = \omega^2 r##, and it's this quantity that enters in on the righthand side of F=ma.
     
  8. Dec 24, 2011 #7
    what forces would be helping or against the rotation of the body?
     
  9. Dec 24, 2011 #8
    consider a case of vertical motion of stone tied to a string ... when the stone is at highest point, the gravity works along with tension but at lowest point gravity works against the tension ... draw free body diagram n you'll understand
     
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