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# Mechanic Question...Need help pleaseee

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1. 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
2. (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?
3. (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
4. (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?
5. (Original post by Zacken)
Can you show me your equation?
6. (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 (from making . So this means that the speed at the bottom, using your energy equation is .

Now resolving radially at the bottom of the circle gets you - solve for T now.
7. (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 (from making . So this means that the speed at the bottom, using your energy equation is .

Now resolving radially at the bottom of the circle gets you - 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
8. (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.
9. (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!
10. (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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