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Homework Help: Whirling stone in vertical circle.

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A stone, attached to one end of an inelastic string whirls around in a vertical circle of 1.0 m radius. The tension in the string when the stone is at the bottom of the circle is found to be four times the tension in the string when the stone is at the top. Find the velocity of the stone when it is at the top of the circle.

    2. Relevant equations
    arad = vt²/r
    vt = rω
    x-direction right = positive
    y-direction up = positive
    assuming it spins at counter clockwise direction = positive

    3. The attempt at a solution
    At the top of the circle:
    -T-mg = marad = -mvtop²/r
    vtop2 = (T+mg)/m

    At the bottom of the circle:
    4T-mg = mvbottom2/r
    vbottom2 = (4T-mg)/m

    At an angle θ counter clockwise from the vertical axis:
    -mgsinθ = mat
    -gsinθ = dVt/dθ x dθ/dt
    ∫ -rgsinθ dθ = ∫ Vt dVt
    rgcosθ = 0.5Vt2 + c

    When θ = 0, and r =1.0, Vt2 = vbottom2 = (4T-mg)/m
    c = g - (4T-mg)/2m
    Therefore, Vt2 = (4T-mg)/m + 2gcosθ - 2g

    When θ = 2π, Vt2 = vtop2 = (T+mg)/m
    (4T-mg)/m - 4g = (T+mg)/m
    T= 2mg

    vtop2 = (T+mg)/m = 3g

    vtop = -(3g)

    Is my answer correct?
  2. jcsd
  3. Apr 4, 2015 #2


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    Homework Helper
    Gold Member

    seems correct to me, though it would ve been a bit more simple if you had use conservation of energy for the top and bottom position.

    Small corection (doesnt affect the final result since it seems you used cos(2pi)=-1), for the top position it is [itex]\theta=\pi[/itex] not [itex]\theta=2\pi[/itex]
    Last edited: Apr 4, 2015
  4. Apr 4, 2015 #3
    Is that an exact reproduction of the problem statement?
  5. Apr 4, 2015 #4
    Thanks, noted my mistake. I was required to do it by forces though.
  6. Apr 4, 2015 #5
    Yes it is the exact question.
  7. Apr 4, 2015 #6
    Then I think your solution is correct.
  8. Apr 4, 2015 #7
    How could you use conservation of energy if the circular motion is not uniform?
  9. Apr 4, 2015 #8


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    Homework Helper
    Gold Member

    Well the work of the tension is zero ( string is inelastic, tension is always perpendicular to velocity), we have only work of the weight involved which is a conservative force so conservation of energy applies.
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