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hahalovex
S.O.S. Newbie
Joined: 17 Oct 2004
Posts: 2
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 Posted: Sun, 17 Oct 2004 00:35:28 UTC Post subject: please help me on the uniform circular motion.. |
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ok..
i got stuck on this question..
Q. A stone of mass 0.85kg is swung in a vertical circle at the end of a string with a fixed speed. if the speed is such that the string becomes just slack at the highest position, what is the tension of the strin gat the lowest position?
I 'm not sure what i have to do first..
the answer was 2mg .
thank you for the help. |
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Matt
Member of the 'S.O.S. Math' Hall of Fame
Joined: 01 Oct 2003
Posts: 8506
Location: Sacramento, CA
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| Posted: Sun, 17 Oct 2004 04:00:33 UTC Post subject: Re: please help me on the uniform circular motion.. |
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| hahalovex wrote: | ok..
i got stuck on this question..
Q. A stone of mass 0.85kg is swung in a vertical circle at the end of a string with a fixed speed. if the speed is such that the string becomes just slack at the highest position, what is the tension of the strin gat the lowest position?
I 'm not sure what i have to do first.. |
Draw force diagrams for the stone at both the top and bottom of the circle. Because the stone is in uniform circular motion, the only force acting on the stone is the centripetal force. That means the net force in both diagrams must be equal to the centripetal force (which points in the direction of the center of the circle).
Consider the force diagram for when the stone is at the top of the circle. Because the string is slack, the only force acting on the stone is the gravitational force. That means the centripetal force is equal to the stone's weight.
Now consider the force diagram for when the stone is at the bottom of the circle. We don't know if the string is slack or not, so we must take the tension in the string into consideration. Now, as I said earlier, the centripetal force is equal to the net force (because the stone is in uniform circular motion). We know the centripetal force is equal to the weight of the stone. If we adopt the convention that upward is the positive direction and downward is the negative direction, then the net force is the tension in the string minus the stone's weight. Hence:
By the way:
| hahalovex wrote: | | the answer was 2mg . |
The answer can't be in milligrams, because tension is measured in Newtons (you should know that!)... |
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CMDP
S.O.S. Oldtimer
Joined: 05 May 2003
Posts: 172
Location: Terneuzen, Netherlands.
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| Posted: Sun, 17 Oct 2004 07:09:17 UTC Post subject: |
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Matthew,
| Quote: | I 'm not sure what i have to do first..
the answer was 2mg .
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2mg is clearly not supposed to be 2 milligrams.....,it is supposed to be 2 times the mass times the acceleration of gravity=2 times the weight of the stone in N,which confirms your answer. |
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Matt
Member of the 'S.O.S. Math' Hall of Fame
Joined: 01 Oct 2003
Posts: 8506
Location: Sacramento, CA
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| Posted: Sun, 17 Oct 2004 08:21:14 UTC Post subject: |
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| CMDP wrote: | | 2mg is clearly not supposed to be 2 milligrams |
'Clearly'? Not clearly to me! 
| CMDP wrote: | | it is supposed to be 2 times the mass times the acceleration of gravity=2 times the weight of the stone in N,which confirms your answer. |
Ah, good observation...but then why would the book provide any numbers (i.e., 0.85kg) at all if the answers are terms of variables?  |
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hahalovex
S.O.S. Newbie
Joined: 17 Oct 2004
Posts: 2
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| Posted: Sun, 17 Oct 2004 11:16:15 UTC Post subject: |
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Thank you very much.
oh, the answer was 2 times mass times gravity.
I'm sorry i didn't make it clear.
My teacher made this problem and ha gave me an answer 2mg.
which means 2 x (weight of the mass)
Maybe he wanted us to know what the real answer, 16.66N, means. 
Again, thank you. 
I think I got the concept of uniform circular motion. =) |
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