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?
m v^2/r
v the same
r the same
twice m
Wait a minute, this horizontal circle does not mean the string is horizontal.
r will not be constant in the real case.
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.
And we already did the limit for very fast, string horizontal also gives twice the tension for twice the mass.
Cute problem !
To understand how the tensions in the strings differ when swinging yo-yos of different masses in a horizontal circle, let's break it down step-by-step.
1. The tension in the string is the force that keeps the yo-yo moving in a circular path. It acts towards the center of the circle and is always directed inward.
2. In circular motion, the tension in the string is responsible for providing the centripetal force required to keep the yo-yo moving in a circular path. This force is given by the equation Fc = (mv^2) / r, where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.
3. When you swing the first yo-yo with a certain mass, the tension in the string is determined by the equation T = m * g, where T is the tension, m is the mass of the yo-yo, and g is the acceleration due to gravity. The tension is equal to the weight of the yo-yo.
4. Now, let's consider the second yo-yo with twice the mass. If we keep the length of the string and period of the yo-yo swing the same, the velocity of the yo-yo will also be the same. The centripetal force required to keep it in circular motion will increase because of the increased mass.
5. According to the equation Fc = (mv^2) / r, the centripetal force is directly proportional to the mass. Therefore, with the second yo-yo having twice the mass, the centripetal force required will also be twice as much.
6. Since the tension in the string is responsible for providing the centripetal force, the tension in the string of the second yo-yo will be twice as much as the tension in the string of the first yo-yo.
In conclusion, when swinging yo-yos of different masses in a horizontal circle with the same length of string and period, the tension in the string of the second yo-yo will be twice as much as the tension in the string of the first yo-yo.