Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cart ride Circular Motion

  1. Oct 1, 2005 #1
    here's a tough one :S

    A 70kg person rides in a 30kg cart moving at 12 m/s at the top of the hill that is the shape of an arc of a circle with radius of 40m.

    a) What is the apparent weight of the person as the cart passes over the top of the hill?

    b) Determine the maximum speed that the cart may travel at the top of the hill without losing contact with the surface. Does this depend on the mass of the cart or that of the person?

  2. jcsd
  3. Oct 2, 2005 #2
    a) We can treat person and cart as an one object. On the top of the hill there are centripetal force:
    F = (v^2 / R) * (m1 + m2)
    And because of the second Newton's law a force -F affects on car with person.
    So the weight of car with person are Fg - F, where Fg is gravity force.
  4. Oct 2, 2005 #3
    So Fg is going to be (m1 + m2)g right?
  5. Oct 2, 2005 #4
    That's correct for the whole cart+person system.

    Ftotal = Fg + Fc = mg - mv^2/r.

    You seem a bit confused by what you should use for 'm.' Here's a good exercise that will help you:

    Solve part a) twice. On the first try, solve for the total force on the cart using mcart then solve for the total force on the person using mperson and add these 2 results together to get Ftotal. On the second try, use m = mperson + mcart. You should get the same answer either way.
  6. Oct 2, 2005 #5
    Ok ok, so after I do that, wouldn't the weight of the person be Ftotal * 0.7 ?

    My answer for the person came out to 434N
  7. Oct 2, 2005 #6
    434N looks correct.

    As for the first question, if by ftotal you mean the total force for the combined cart-person system, then yes, that is correct. Since v, r, and g are the same for the person and the cart, the total force on the cart and the person is proportional their masses.
  8. Oct 2, 2005 #7
    Cool, what about b) ? I don't think I can count in friction here?
  9. Oct 2, 2005 #8
    We're neglecting friction here. If you're referring to ftotal above, that should be an 'F' for force, not friction.

    For part b), you've got 2 forces acting on the cart/person. One is gravity, Fg=mg, pulling downward. The other is the centrigugal 'force', Fc=mv^2/r pulling downwards. When these 2 forces are equal, the cart/person will be weightless. Try setting Fg equal to Fc and see if you can solve for velocity.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook