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    A stone of mass 0.5kg performs complete revolutions in a vertical circle on the end of a light, inextensible string of length 1m. Show that the string must be strong enough to support a tension of at least 29.4.

    I can make equation of the KE and GPE of the stone when it is at the bottom and the top and then make the equation of 'Tension+mg=ma' ... I don't know whether these are the right steps or not...Can't get the answer...need help plzzzz
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    (Original post by Oliviazh)
    A stone of mass 0.5kg performs complete revolutions in a vertical circle on the end of a light, inextensible string of length 1m. Show that the string must be strong enough to support a tension of at least 29.4.

    I can make equation of the KE and GPE of the stone when it is at the bottom and the top and then make the equation of 'Tension+mg=ma' ... I don't know whether these are the right steps or not...Can't get the answer...need help plzzzz
    You're doing fine for the moment! What seems to be the trouble?
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    (Original post by Zacken)
    You're doing fine for the moment! What seems to be the trouble?
    Well, I can make the equations but can't solve them...it's a negative number which doesn't make any sense...There must be some mistakes in my equations I think
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    (Original post by Oliviazh)
    Well, I can make the equations but can't solve them...it's a negative number which doesn't make any sense...There must be some mistakes in my equations I think
    Can you show me your equation?
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    (Original post by Zacken)
    Can you show me your equation?
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    (Original post by Oliviazh)
    ...
    Okay, so I think you need to be considering the bottom instead - actually. You know that at the top, the minimum speed is \sqrt{g} (from making \frac{v^2}{2} = \frac{g}{2} \Rightarrow v^2 = 2. So this means that the speed at the bottom, using your energy equation is  g + 39.2 = u^2 \Rightarrow u^2   = 49 .

    Now resolving radially at the bottom of the circle gets you T - mg = \frac{u^2}{2} - solve for T now.
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    (Original post by Zacken)
    Okay, so I think you need to be considering the bottom instead - actually. You know that at the top, the minimum speed is \sqrt{g} (from making \frac{v^2}{2} = \frac{g}{2} \Rightarrow v^2 = 2. So this means that the speed at the bottom, using your energy equation is  g + 39.2 = u^2 \Rightarrow u^2   = 49 .

    Now resolving radially at the bottom of the circle gets you T - mg = \frac{u^2}{2} - solve for T now.
    Um...well...I understand the last part...but I don't know how you get the min speed at the top
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    (Original post by Oliviazh)
    Um...well...I understand the last part...but I don't know how you get the min speed at the top
    Minimum speed occurs when the centripetal force equals the weight.
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    (Original post by Zacken)
    Minimum speed occurs when the centripetal force equals the weight.
    Oh... I understand now...I've got the right answer...Thank you so much!
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    (Original post by Oliviazh)
    Oh... I understand now...I've got the right answer...Thank you so much!
    You're very welcome.
 
 
 
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