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Math Help - Vertical circular motion - Mechanics 2

  1. #1
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    Vertical circular motion - Mechanics 2

    Small bead mass denoted by m, suspended from fixed point O by a light inextensible string length a. Then set into motion with the string taut at B where B is vertically below O which horizontal speed u.

    Given that the string does not become slack, show the least value of u required for the bead to make complete revolutions about O is 5ag^1/2.

    I have it to where u^2=v^2 +4ag through using mechanical energy principle but I fail to see how v^2 =ag when v is greater than or equal to 0 for complete revolutions where v is the highest point of the circle.

    On the mark shceme they aquired v^2 =ag by stating (mv^2)/a=mg but I thought it should surely be (mv^2)/a=T +mg especially considering the string never becomes slack.

    Help please...
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  2. #2
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    Re: Vertical circular motion - Mechanics 2

    Quote Originally Posted by packer View Post
    Small bead mass denoted by m, suspended from fixed point O by a light inextensible string length a. Then set into motion with the string taut at B where B is vertically below O which horizontal speed u.

    Given that the string does not become slack, show the least value of u required for the bead to make complete revolutions about O is 5ag^1/2.

    I have it to where u^2=v^2 +4ag through using mechanical energy principle but I fail to see how v^2 =ag when v is greater than or equal to 0 for complete revolutions where v is the highest point of the circle.

    On the mark shceme they aquired v^2 =ag by stating (mv^2)/a=mg but I thought it should surely be (mv^2)/a=T +mg especially considering the string never becomes slack.

    Help please...
    Let the velocity of the bead at the top of the circle be $v_{top}$

    In order to keep the string taut centripetal force must be at least that of gravity.

    $m \dfrac {v_{top}^2}{a} \geq m g$

    $v_{top}^2 \geq a g$

    Now at the bottom the bead has kinetic energy

    $K=\dfrac 1 2 m u^2$

    To get the to top of the circle it trades

    $P=2m a g $ in kinetic energy for potential energy.

    So

    $K_{top}=\dfrac 1 2 m v_{top}^2 = \dfrac 1 2 m u^2 - 2 m a g$

    $v_{top}^2 = u^2 - 4 a g \geq a g$

    $u^2 \geq 5 a g$

    $u \geq \sqrt{5 a g}$
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    Re: Vertical circular motion - Mechanics 2

    Thankyou for the help!
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