Posted by **David** on Wednesday, April 21, 2010 at 7:24pm.

You swing one yo-yo around your head in a horizontal circle. Then you swing another yo-yo with twice the mass of the first one, but you don't change the length of the string or the period. How do the tensions in the strings differ?

- Physics -
**Damon**, Wednesday, April 21, 2010 at 7:30pm
m v^2/r

v the same

r the same

twice m

- Physics -
**Damon**, Wednesday, April 21, 2010 at 7:32pm
Wait a minute, this horizontal circle does not mean the string is horizontal.

r will not be constant in the real case.

- Physics -
**Damon**, Wednesday, April 21, 2010 at 7:47pm
Say string at angle A from vertical.

Then r of circle = R sin A if R is string length

Inward force on mass by string = T sin A

so

T sin A = m v^2/r = m v^2/(R sin A)

T sin^2 A = m v^2/R

Vertical force on mass by string = T cos A

so

T cos A = m g

period = p = 2 pi r/v = 2 pi R sin A/v

so

v = 2 pi R sin A/p

and

v^2 = 4 pi^2 R^2 sin^2A/p^2

so

T sin^2 A = m v^2/R = m * 4 pi^2 R^2 sin^2A/p^2

so

T = m 4 pi^2 R^2/p^2

interesting, if p is the same when you double m you double the tension

I suppose that makes sense in the limit of very slow and hanging just about straight down, the tension is mostly m g and has to be double for twice the mass.

- Physics -
**Damon**, Wednesday, April 21, 2010 at 7:58pm
And we already did the limit for very fast, string horizontal also gives twice the tension for twice the mass.

Cute problem !

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