Analyzing the Tension of a Yo-Yo String

In summary, 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, the tension in the string must exist with a value of 6N.
  • #1
bilbobaggins
12
0

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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  • #2
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.
 
  • #3
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
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
 
  • #5
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.
 
  • #6
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?
 
  • #7
I would say that's right, for a horizontal circle.
 
  • #8
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?
 
  • #9
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?
 

What is the tension of a yo-yo string?

The tension of a yo-yo string refers to the force or tightness of the string when the yo-yo is in motion. It is the force that keeps the yo-yo attached to the string and allows it to move up and down.

How is the tension of a yo-yo string measured?

The tension of a yo-yo string can be measured in units of force such as pounds or newtons. It can also be measured indirectly by observing the length of the string when it is at rest and comparing it to the length when the yo-yo is in motion.

What factors affect the tension of a yo-yo string?

The tension of a yo-yo string can be affected by various factors such as the material and thickness of the string, the weight and design of the yo-yo, and the speed and direction of the yo-yo's motion. The tension can also be affected by external forces such as gravity and air resistance.

How does the tension of a yo-yo string affect its performance?

The tension of a yo-yo string plays a crucial role in its performance. A higher tension can increase the yo-yo's responsiveness, making it easier to control and perform tricks. However, too much tension can also cause the yo-yo to bind or snag, hindering its movement.

How can the tension of a yo-yo string be adjusted?

The tension of a yo-yo string can be adjusted by changing the length of the string or by altering the weight and design of the yo-yo. Some yo-yos also have adjustable mechanisms or bearings that can affect the tension. Additionally, proper maintenance and care of the yo-yo and string can help maintain optimal tension.

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