A yo-yo has a string length of 0.200 m

What is the slowest speed at which you can spin it to keep it in a fixed path?

What is this, and how do you find it?
Thanks

define "fixed path"

The question just says "the slowest speed at which you can spin it to keep it in a fixed path"

Ok, then I can't help, a fixed path is a straight line. I can't imagine that with a yo-yo. So if it is in a circle, is it horizontal circle or vertical. Normally a yoyo would be operating in a vertical orbit, but that hardly is a fixed path. I will see if I can get you some other eyes on this.

Im sorry, I forgot...

Its a vertical circle.

If it is a vertical path then and it is sleeping at the end of the string, it is like a rock at the end of a string and the centripetal acceleration at the top of the loop must be equal or greater than 1 g or the string will go slack and the rock fall out of the vertical circle.

v^2/r >/= g for circular path
so v at least sqrt ( g r )
v = sqrt (9.8 * .2) = sqrt (1.96)
v = 1.4 m/s

To find the slowest speed at which you can spin the yo-yo to keep it in a fixed path, we need to determine the minimum speed required to balance the gravitational force acting on the yo-yo. This occurs when the centripetal force directed towards the center of the circular path is equal to the gravitational force.

To calculate this speed, we can use the following equation:

v = √(g * r)

where:
v is the minimum speed required to keep the yo-yo in a fixed path,
g is the acceleration due to gravity (approximately 9.8 m/s²),
and r is the radius of the circular path (which is half the length of the string).

Given that the string length is 0.200 m, the radius of the circular path is 0.200 m / 2 = 0.100 m.

Now we can plug these values into the equation to find the minimum speed:

v = √(9.8 m/s² * 0.100 m) = √(0.98 m²/s²) ≈ 0.99 m/s

Therefore, the slowest speed at which you can spin the yo-yo to keep it in a fixed path is approximately 0.99 m/s.