Analyzing the Tension of a Yo-Yo String

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Homework Help Overview

The problem involves analyzing the tension in a yo-yo string as the yo-yo makes a complete circular revolution. The subject area pertains to circular motion and forces acting on the yo-yo.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss drawing diagrams and labeling forces, finding the velocity of the yo-yo, and applying Newton's second law in the context of circular motion. There are attempts to calculate centripetal force and acceleration, with some questioning the assumptions about the direction of the motion.

Discussion Status

Participants are exploring various interpretations of the problem, particularly regarding the assumptions about the motion being horizontal or vertical. Some guidance has been offered regarding the application of equations, but there is no explicit consensus on the correct approach.

Contextual Notes

There is uncertainty about whether the yo-yo is moving in a horizontal or vertical circle, which affects the analysis of forces involved. Participants also express confusion about the implications of changing the tension in the string.

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Homework Statement




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.

Homework Equations


F+t=ma
fc equations
a=f/m

The Attempt at a Solution


they're so wrong, they wouldn't help.
 
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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.
 
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.
 
hage567 said:
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 here's 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 that's wrong
 
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?
bilbobaggins said:
okay here's 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 that's wrong

Note: I am guessing you are saying the yo-yo is traveling 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.
 
hage567 said:
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 traveling 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?
 
I would say that's right, for a horizontal circle.
 
hage567 said:
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?
 
I'm not sure what you mean by double the tension. Double it by doing what?
Look at the equation:
[tex]T=\frac{mv^2}{r}[/tex]
If T was replaced by 2T, what would that do to v?
 

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