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    • Thread Starter


    What I've done so far:

    Using conservation of momentum:
    mu = (m+M)V
    V = mu/(m+M)

    Change in kinetic energy:
    1/2mu² - 1/2(m+M)(mu/(m+M)) = 1/2mu² - 1/2mu = 1/2mu(u-1)

    For the next part: I assumed the equation of motion would be:
    T + (m+M)g = [(m+M)V²]/R

    But does the velocity remain constant?

    (Original post by SunGod87)
    Change in kinetic energy:
    1/2mu² - 1/2(m+M)(mu/(m+M)) = 1/2mu² - 1/2mu = 1/2mu(u-1)
    You forgot to square that.

    Now, the velocity certainly doesn't remain constant as the particle is changing direction. So let's find the velocity v when the particle is at the top using conservation of energy:
    0.5(m+M)V^2 = (m+M)g(2R) + 0.5(m+M)v^2
    => v^2 = V^2 - 4gR

    So we have:
    T = (m+M)v^2/R - (m+M)g
    = (m+M)(v^2 - gR)/R
    = (m+M)(V^2 - 5gR)/R, as required.

    The string remains taught as long as T>=0, that is, as long as:
    V^2 >= 5gR
    (mu/(m+M))^2 >= 5gR

    So the minimum value of u is:
    (m+M)/m * sqrt(5gR)

    And finally:
    T_u - (m+M)g = (m+M)V^2/R
    = (m+M)(5gR)/R
    = 5(m+M)g
    => T_u = 6(m+M)g
    • Thread Starter

    Thanks very much

    For some reason I was assuming at the top all of the energy was potential and none kinetic, silly me.
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Updated: April 11, 2006
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