# Tension of a yo-yo string?

1. May 21, 2007

### bilbobaggins

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.

2. May 21, 2007

### hage567

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.

3. May 21, 2007

### hage567

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.

4. May 21, 2007

### bilbobaggins

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

5. May 21, 2007

### hage567

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.

6. May 21, 2007

### bilbobaggins

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?

7. May 21, 2007

### hage567

I would say that's right, for a horizontal circle.

8. May 21, 2007

### bilbobaggins

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?

9. May 21, 2007

### hage567

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?