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Circular Motion of car over a bridge

  1. Aug 21, 2007 #1
    I. A hump back bridge has a radius of curvature of 40m. Calculate the max speed at which a car could travel across the bridge without leaving the road at the top of the hump.

    I realised that weight - contact force at the top would give the resultant centripetal force.

    But i have no equation for the contact force.

    II. Calculate the speed at which a plane must be flying when looping-the-loop of radius 0.80 km so that the pilot feels no force from eitherhis harness or his seat.

    component of weight towards the centre - contact force = resultant centripetal force

    Here again i have no equation for the contact force.

    I'm wondering if i should be putting the conatct force to zero??
  2. jcsd
  3. Aug 21, 2007 #2
    No...i'm mistaken...weight will depend on the position in the vertical circle.

    someone please clear this doubt.....

    thank you
  4. Aug 21, 2007 #3


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    I. Well, think how the mass cancels out when you equate the gavitational force and the centripetal force [tex] F_{centripetal} = m \frac{v^2}{r} [/tex]. By the way, I assumed by radius of curvature, you meant the radius of the semi-circular curve formed by the bridge.

    II. If he is feeling no force, the pilot must be at the bottom of the loop where the centripetal force just balances the gravitational force, both questions are the same, conceptually.
  5. Aug 22, 2007 #4
    Two more questions:

    III. A particle is moving in a circular path described by the equation:
    'theta'= (3 rad s-2t^2) - (2 rad s-1)t

    Calculate the angular velocity and the angular acceleration at t = 6.0s.

    I'm tootally lost.

    IV. A particle is slightly displaced from rest at the top of a fixed smooth sphere of radius r. Find the vertical height through which the particle descends before leaving the sphere.

    I realised that the compponent of weight - contact force = resultant centripetal.

    I have no clue about what should be the contact force as in the above questions.
  6. Aug 22, 2007 #5


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    They're asking for the maximum speed... suppose there's a frictional force acclerating the car forward (frcitional force acting on the tires is what pushes the car forward), then there'd be a normal force... hence that would make weight-contact force less... hence the speed would be less for centripetal acceleration...

    So assume the contact force is 0 for the maximum speed(this means that there's no frictional force acting, so the instantaneous tangential acceleration of the car is 0... all the accleration is centripetal)

    This is at the top of the loop... the force acting on the pilot is mg... so if the harness or seat weren't there he'd accelerated downward by g... the harness and seat (which are part of the plane) are accelerating downward by v^2/r. If v^2/r > g then the pilot would feel the seat pushing against him... if v^2/r<g then the pilot is accelerating faster than the harness, so he'd get caught in it and feel the harness...

    so you want v^2/r = g.
  7. Aug 22, 2007 #6


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    The rads^-2 and rads^-1 are just units... I'd write theta as: 3t^2 - 2t

    So calculate [tex]\frac{d\theta}{dt}[/tex] which is angular velocity and calculate: [tex]\frac{d^2\theta}{dt}[/tex] which is angular acceleration

    The way I think about this is... when a particle is moving in a circular path like this...
    if v^2/r is less than the acceleration towards the center due to gravity, then the particle will want to go inwards, towards the center of the circle... but the normal force prevents this from happening... when v^2/r is greater than the acceleration towards the center due to gravity... the particle will want to outwards, away from the center... so there's no normal force, and nothing to stop the particle from going off the circular path.

    You can use this reasoning for the car/bridge problem also... It is exactly when the car hits the maximum possible speed, that the normal force becomes 0. v^2/r = g is the moment this happens.

    I see the normal force in these problems as a response to whatever speed the object is moving at... So you know 2 things... at the maximum speed the normal force is 0 and v^2/r equals the centripetal acceleration due to all the other forces (other than the normal force)...

    In this particular problem, you can use conservation of energy (gravitational potential energy and kinetic energy) to find v^2/R... and you can also find the acceleration towards the center due to gravity which is in terms of [tex]\theta[/tex]. the angle when v^2/R = acceleration towards center due to gravity is your answer.
  8. Aug 23, 2007 #7
    i have yet another question.... when you are swirling smething attached to a string and doing circular motion. the weight is due to the mass of the body. But what causes the tension. the centripetal force is the resultant between the tension and weight. and what does the centrifugal force come to do in that?
  9. Aug 23, 2007 #8
    another thing...my teacher told me that the centripetal force is a resultant force. considering question II, what is actually the force towards the centre of the loop, which gives a resultant force with weight, the centripetal force. i understaood what you meant 'learning physics' for this question, except for the above stuff. thnks a lot. thnks to bel too.
  10. Aug 23, 2007 #9


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    Your hand.

    If you take the rotating frame as the rest frame... let's say you're in some enclosed room which is being rotated circularly... if you're inside that room, you will feel like you're being pushed outward... we say that the centrifugal force is acting on you.... in reality the room is being pushed inward... if you measure accelerations of objects inside the room, relative to the room (ie the accelerations you'd measure if you were at rest in the room)... and multiply by the mass of those objects... that gives the centrifugal force acting on those objects.
  11. Aug 23, 2007 #10


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    For the pilot at the top of the loop, they said he feels nothing... that means there's 0 contact force from the seat or harness. Hence, the centripetal force is exactly his weight. Weight is the only force acting on the pilot.
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