Tension of a yo-yo string?

Question

The yo-yo has a mass of .2kg and is attached to a sting .8 m long. If the yo-yo makes a complete circular revolution each second, what tension must exist in the string? Can anyone please show me how to do this. thanks.

Answer 1

A couple of questions:
Is the yo-yo descending, ascending, or "sleeping" at the end of the string?

If sleeping, then T=m*g
Where T is the tension.

If the yo-yo is descending or ascending then
start with translational:

I will assume the string plays out at radius r from the center, which can be different than R, the radius of the yo-yo
m*g-T=m*a
where T is the tension and acceleration is the acceleration of the yo-yo. Sign for a will matter depending on the direction of motion.
isolate a
g-T/m=a

Next rotational:
T*r=I*a/R
I used the alpha=a/R to get one a
Isolate a
T*R*r/I=a

now
T*R*r/I=g-T/m

simplify


T*(1/m+R*r/I)=g

The only unknown is I, the moment of inertia for the yo-yo.

If you model it as a disk with the string playing out at the radius, then R=r and I=.5*m*R^2

substituting that in
T=g*m/3
if the string is playing out from a center axle, then r is not=R,
but I is still a pretty good approximation to .5*m*R^2, as long as the rod mass is negligible. If the rod mass is not negligible, it is likely that the inertia of the axle is since the I for the axle is .5*m(axle)*r^2, and r< Under these conditions

T=g*m*R/(R+2*r)

BTW: if the yo-yo is ascending, then
g=T*(1/m-r*R/I)
j

Answer 2

that is what i was going to say