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bilbobaggins
May21-07, 06:19 PM
1. The problem statement, all variables and given/known data


The yo-yo has a mass of .2kg and is attached to a sting .8 m long. If the yo-yo makes a complete circular revolution each second, what tension must exist in the string? Can anyone please show me how to do this. thanks.
2. Relevant equations
F+t=ma
fc equations
a=f/m

3. The attempt at a solution
they're so wrong, they wouldn't help.

hage567
May21-07, 06:40 PM
Have you drawn a diagram and labelled all the forces acting on the yo-yo? What do you know about circular motion? What information can you get from the question?

they're so wrong, they wouldn't help.

Not true. They could help us figure out where you're going wrong.

hage567
May21-07, 06:47 PM
How about trying to find the velocity of the yo-yo as it goes around the circle? You can get that from the info in the question. Give it a try.

bilbobaggins
May21-07, 06:51 PM
Have you drawn a diagram and labelled all the forces acting on the yo-yo? What do you know about circular motion? What information can you get from the question?



Not true. They could help us figure out where you're going wrong.

okay heres what i did, i got the circumference was 15.77 from going 2 *.8 * 3.14 = 5.024. then it takes one second to go around the circle so velocity is 5.024m/s

then use fc= m*v/r
i got fc = 49.73

then i did a=f/m
49.73N/.2kg

248.65 = a

F+T= ma
49.73 +T= .2kg * 248.65
so t = 6

But I think thats wrong

hage567
May21-07, 07:26 PM
The tension in the string is acting as the central force. So you must use Newton's second law in the form for circular motion. Are we assuming the yo-yo is in a horizontal circle?
okay heres what i did, i got the circumference was 15.77 from going 2 *.8 * 3.14 = 5.024. then it takes one second to go around the circle so velocity is 5.024m/s This looks OK.

then use fc= m*v/r
The v term should be squared. This is really Newton's second law, with centripetal acceleration in place of linear acceleraton.
i got fc = 49.73

then i did a=f/m
49.73N/.2kg
This is not correct. You've already found the centripetal acceleration above, that is what you need. This equation is for linear motion.

248.65 = a

F+T= ma
49.73 +T= .2kg * 248.65
so t = 6
There is only one force acting on the yo-yo, and that is the tension in the string.
But I think thats wrong

Note: I am guessing you are saying the yo-yo is travelling in a horizontal circle. If it was a vertical circle, we would have to take gravity into account as well. So I'm not sure which way it should be done.

bilbobaggins
May21-07, 07:31 PM
The tension in the string is acting as the central force. So you must use Newton's second law in the form for circular motion. Are we assuming the yo-yo is in a horizontal circle?


Note: I am guessing you are saying the yo-yo is travelling in a horizontal circle. If it was a vertical circle, we would have to take gravity into account as well. So I'm not sure which way it should be done.

okay so is it 6.3? Equation used

.2*5.024 m/s ^2/ .8m = 6.31
is that as far as i have to go?

hage567
May21-07, 07:37 PM
I would say that's right, for a horizontal circle.

bilbobaggins
May21-07, 07:43 PM
I would say that's right, for a horizontal circle.

ok, thank you for the help. Oh and one more question, If I doubled the tension, would speed change? I don't think it would would it?

hage567
May21-07, 07:53 PM
I'm not sure what you mean by double the tension. Double it by doing what?
Look at the equation:
T=\frac{mv^2}{r}
If T was replaced by 2T, what would that do to v?