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Physics "Circular Loop" question.

Discussion in 'MCAT Study Question Q&A' started by dorjiako, Aug 13, 2011.

  1. dorjiako

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    Physics "Circular Loop" question.
    The initial question is this: A 1kg block slides down a ramp and then around a circular loop of radius 10m. Assuming that all surfaces are frictionless, what is the minimum height of the ramp so that the block makes it all the way around the loop without falling.
    Then, these questions were asked.
    1. What is the normal force at the bottom of the loop if the height of the ramp is four times that of the loop?
    2. How fast does the block need to be going at the bottom of the ramp so that the acceleration of the block at the top of the loop is 4g?
    3. What is the speed of the block as it exits the loop if the normal force at the top of the loop was 80N?
    Thanks for helping me out with these questions.
     
  2. costales

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    Minimum speed at top is when N = 0 and the net centripetal force is mg = mv^2/r, giving v at top. Energy at release height h must equal energy at top: mgh = mg(2r) + 0.5mv^2 where v is speed you found. Now you can solve for h.
    1. If released from 8r, speed at bottom is calculated using energy conservation: mg(8r) = 0.5mv^2, use this speed to get N at bottom from the net centripetal force N - mg = mv^2/r.
    2. Centripetal accel at top is 4g = mv^2/r, solve for v at top. Use this to solve for speed at bottom v' in energy equation 0.5mv'^2 = mg(2r) + 0.5mv^2.
    3. Use N + mg = mv^2/r and solve for v at top. From this height 2r, use energy conservation to get v' at bottom.
     
  3. Majik

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    1.) Consider the forces acting at the bottom of the loop:

    a. Normal Force points Upwards
    b. Gravity Points Downwards

    The net force (centripetal force) also points upwards, therefore:
    mv^2/r = N - mg
    mv^2/r + mg = N

    The velocity here is the max velocity at the bottom of the ramp. The radius is the radius of the loop. The mass is the mass of the block. We can plug in the values they provided to find the normal force, but first using conservation of mechanical energy, let's find the final velocity.

    Initially we have PE only. At the bottom of the ramp, all that energy is converted entirely to KE. (We're told there's no friction).

    mgh = 1/2mv^2
    Solving for v, we get: v = sqrt(2gh); v = sqrt(2x10x80); v = sqrt(1600); v = 40m/s
    Note: The height of the loop is 2x the radius: 20m. The height of the ramp is 4 times the height of the loop, 80m.

    Plugging in this velocity to find Normal Force, we get:
    (1kg)(40m/s)^2/(10m) + (1kg)(10m/s^2) = N
    160N + 10N = 170N
    Normal Force = 170N

    2.) Centripetal Acceleration = v^2/r. We can solve for tangetial velocity and use conservation of ME. 4g = 40m/s^2. 40m/s^2 x 10m = 400 = v^2. sqrt(400) = v; v=20m/s

    Therefore, KE at the top is: 1/2(1)(20)^2 = 200J. PE at the top is: (1)(10)(20) = 200J
    Total ME at the top of the loop is: 400J. Final KE is: 400J = 1/2mv^2; sqrt(800) = v=~28m/s

    3.) Consider the forces at the top of the loop:
    a. Normal Force points Downwards
    b. Gravity points Downwards

    Centripetal Force (net force) also points down. Therefore: mv^2/r = N + mg
    Plugging in the appropriate variables and solving for mv^2, we get: (80N + 10N)10
    900N = mv^2 divide this by 2 to get: 1/2mv^2 (KE)
    450J = KE at the top of the loop.

    PE at the top of the loop equals: mgh (1x10x20m) = 200J
    Therefore, using conservation of ME, the final velocity equals:

    PEi + KEi = KEf (note: PEf = 0J since we're at the bottom of the loop)
    650J = KEf
    650J = 1/2mv^2
    1300J / 1kg = v^2
    sqrt(1300) = v
    ~36 m/s
     
  4. dorjiako

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    I truly appreciates your help. Thanks.
     

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