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Circular Motion and Work

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    An amusement park ride consists of a ring of radius A from which hang ropes of length l with seats for the riders as shown in Figure I. When the ring is rotating at a constant angular velocity, omega, each rope forms a constant angle, theta, with the vertical as shown in Figure II. Let the mass of each rider be m and neglect friction, air resistance, and the mass of the ring, ropes, and seats.

    (A picture is given here on page five: www.swcp.com/~gants/calendars/ap%20physics/sept%20docs/(A)%20Newton's%20LawsC.doc)

    Determine the minimum work that the motor that powers the ride would have to perform to bring the system from rest to the constant rotating condition of Figure II. Express your answer in terms of m, g, l, theta, and the speed v of each rider.

    2. Relevant equations
    Work=change in kinetic energy
    work=force*displacement
    force=m*a(centripetal)

    3. The attempt at a solution
    I know that the force causing centripetal motion is tension.
    F(net, x) = m*a(centripetal) = T * sin (theta)
    F(net, y) = T*cos(theta) - m*g
    T = m*g*sec(theta)
    force = m*a(centripetal) = m*g*tan(theta)
    displacement = L * sin(theta)
    W = force*displacement = (m*g*tan(theta))*(l*sin(theta))

    delta(KE)=1/2*m*(v(final))^2-1/2*m*(v(initial))^2, since v(initial)=0,
    W = delta(KE) = 1/2*m*(v(final))^2

    So I have two separate equations for work. Have I done something incorrectly?
     
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2

    Doc Al

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    Staff: Mentor

    Don't attempt to calculate the work directly by analyzing forces. (Note that the centripetal force that you used is only the final centripetal force. And that there must be a tangential force while the ride is increasing speed.)

    Instead, just figure out the total change in energy of each rider.
     
  4. Nov 4, 2007 #3
    Does that mean that the work is just 1/2*m*v^2?
     
  5. Nov 4, 2007 #4

    Doc Al

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    Staff: Mentor

    No. Consider the total energy.
     
  6. Nov 4, 2007 #5
    I'm sorry, I'm just getting more confused. Does this involve integration?
    work = integral (F ds)

    Fnet = m/r (dx/dt)^2

    And when you talk about total energy, so far I have only looked at kinetic. If I look at potential energy, is that relevant?
     
  7. Nov 4, 2007 #6

    Doc Al

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    Potential energy is most definitely relevant. Total mechanical energy equals kinetic plus potential.
     
  8. Nov 4, 2007 #7
    K1 + P1 = K2 + P2 (mechanical energy remains constant)
    K1 = 0
    P1 = K2 + P2

    So this means that the rotating figure must have less potential energy than the still one. Where is the potential energy from? It can't be gravity because the second one should have a higher potential energy due to gravity because it's higher off the ground.

    And does the solution involve integration of force at all?

    :confused:
     
  9. Nov 4, 2007 #8

    Doc Al

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    What makes you think mechanical energy is constant? A motor is required to get this thing moving!
    The only potential energy in this problem is gravitational.

    Nope. You'll smack yourself when you realize how easy it is.
     
  10. Nov 4, 2007 #9
    umm... K2 +P2 = 1/2*m*v^2 + (L-L*cos(theta))*m*g ??

    where L-L*cos(theta) is the change in height.
     
  11. Nov 4, 2007 #10

    Doc Al

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    Staff: Mentor

    You got it.
     
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