Posted by David on Wednesday, April 21, 2010 at 7:24pm.
You swing one yoyo around your head in a horizontal circle. Then you swing another yoyo 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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