# Physics

posted by on .

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 - ,

m v^2/r
v the same
r the same
twice m

• Physics - ,

Wait a minute, this horizontal circle does not mean the string is horizontal.
r will not be constant in the real case.

• Physics - ,

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 - ,

And we already did the limit for very fast, string horizontal also gives twice the tension for twice the mass.
Cute problem !