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Mechanic Question...Need help pleaseee

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:frown:...need help plzzzz:smile:
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Zacken
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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:frown:...need help plzzzz:smile:


You're doing fine for the moment! What seems to be the trouble?
Reply 2
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Oliviazh
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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
Reply 3
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Zacken
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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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Oliviazh
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Original post by Zacken
Can you show me your equation?

image.jpg
Reply 5
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Zacken
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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 g\sqrt{g} (from making v22=g2⇒v2=2\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=u2⇒u2=49 g + 39.2 = u^2 \Rightarrow u^2 = 49 .

Now resolving radially at the bottom of the circle gets you T−mg=u22T - mg = \frac{u^2}{2} - solve for T now.
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Oliviazh
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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 g\sqrt{g} (from making v22=g2⇒v2=2\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=u2⇒u2=49 g + 39.2 = u^2 \Rightarrow u^2 = 49 .

Now resolving radially at the bottom of the circle gets you T−mg=u22T - 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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Zacken
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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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Oliviazh
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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!:h:
Reply 9
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Zacken
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Original post by Oliviazh
Oh... I understand now...I've got the right answer...Thank you so much!:h:


You're very welcome. :h:

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