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Circular Motion - Newton's Laws in different reference frames

  1. Jul 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A child stands near the middle of a roundabout that is rotating with some angular velocity w. The child moves towards the edge of the roundabout in a straight line from the child's perspective and at constant speed. Explain in as much detail as you can (and using equations) the motion of the child from both the child's perspective and the inertial reference frame. Use Newton's Laws in both cases and comment on the need to introduce fictitious forces in the rotating reference frame.

    2. Relevant equations
    a = v^2/r F = mv^2/r

    3. The attempt at a solution
    Inertial reference frame - sees child's motion as a spiral. Forces would be centripetal force and a tangential force both of which are friction. As the child steps further out their speed must increase because all points along the line they walk must complete a cycle in the same time so the further out you go the bigger the circumference and the higher the speed needs to be. This is provided by a frictional force as they step further out which acts along the tangent. The centripetal force is also fraction towards the centre.

    Rotating frame - child sees they are walking in a straight line. Must still exert a frictional force tangent to the path as they move out and towards the centre the whole time. Newton's laws fail here because the child is walking in a straight line at constant speed yet has to exert two resultant forces in order to do this. Coriolis and centrifugal forces must be introduced otherwise Newton's laws would fail.

    Bits I am stuck with:
    How does the centripetal acceleration change as you go near the edge? At first I though that since the velocity is increasing then so must the centripetal acceleration but the radius is not constant so I am unsure how exactly the centripetal force changes. The radius increases which would decrease the centripetal acceleration but the velocity is increasing.

    Something also seems paradoxical to me so I must be missing something. If you are near the middle of a roundabout then to walk to the edge you must have a resultant force towards the edge otherwise you couldn't move. However, the roundabout is constantly in motion so the resultant force must always be to the centre? What am I missing?

    Thanks in advance
     
  2. jcsd
  3. Jul 12, 2017 #2

    Merlin3189

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    You have an equation for the centripetal force, but it involves v and r which are varying. Perhaps if you used a different formula which had only one of these varying quantities, that would help. Hint: what variable stays constant while the child stays on the same radius (or diameter?) (The roundabout is rotating.)

    When you walk downhill, which way do you apply force?
     
  4. Jul 12, 2017 #3
    Thanks. I'm not sure which equation I could use? If you walk downhill you apply a force uphill. If your walking radially outwards you would apply a force of friction radially inwards onto the floor, the floor would push back and provide a resultant force radially outwards onto you. This is why I'm confused. The centripetal force is radially inwards exerted on the person and it is there all the time. How can you get a radially outward resultant force on the person (which you need to walk radially outward) when, to go round in a circle, you always have to have a radially inward resultant force.
     
  5. Jul 12, 2017 #4

    Merlin3189

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    Equation: you quote ## F = \frac {mv^2} {r} ## but both v and r change as they walk out along the radius.
    Do you know formulae in terms of the angular velocity, ω ?
    AngularMeasure.png HillAnalogy.png
    If you don't already know the formula, just substitute rω for v in your equation.
    On a spinning disc, points along the radius have different speeds, but they all move through the same angle in the same time, so they have the same angular speed ω. So you can get F as a function of r and you can see how the force varies with r.

    Centripetal force: At any point except the centre, you need to provide a centripetal force towards the centre to stay in the same place on the disc.
    If you stood on top of a dome, similar conditions apply. Considering only the radial force component, not any vertical component, you need no radial force to stand on the top. At any other point you need to provide a centripetal force to stop yourself sliding outward (and downward).
    If you want to move radially outwards on the dome (down the slope) what would you need to do to move at constant speed?

    NB. There are differences. On the dome the radial force will not vary in the same way. On the rotating disc, as well as the vertical and radial forces, you may need to consider other directions.

    This appears to be a question to make you think, rather than calculate values, so I'm trying not to tell you what to think. I suggest angular measure as a solution to your problem about understanding a formula where two quantities vary at once. I suggest the dome as an analogy to help you see how it might feel. You need to explain your ideas about the original problem and we can point out any errors and limitations to help you develop them.
     
  6. Jul 12, 2017 #5

    Merlin3189

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    So why do you move in the opposite direction to that force?

    If I stand still on the side of a hill, I need to apply a certain force uphill just to stay still. What do I do now to make myself move downhill?
    Think about the size of the force.
    In many situations there are several forces acting on a body. What happens depends on the balance of forces or the resultant force. Standing on the floor, gravity is pulling me down and the floor is pushing me up (I hope!). If they are equal, there is no resultant and I stay where I am. If I want to jump up, I must push harder on the floor, so that it will push harder on me (Newton 3rd). Then it will be a greater force than gravity, the resultant will be up and I will rise (for a very brief moment until my feet are no longer pushing on the floor.) I could also lower my CG by pushing less hard on the floor (relaxing leg muscles) so that gravity is winning for a moment and I slump to the floor.
     
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