Answer:

Step-by-step explanation:
For this case we can use the following formula for the angular velocity:

where
represent the final angular velocity ,
the initial angular velocity , t the time and
the angular acceleration.
And for the linear acceleration we have this formula:

Where a represent the linear acceleration and
the angular acceleration.
For this case the linear acceeleration is given

And the radius of the yo-yo is also given

So then we can use the following formula:

If we replace the values we got:

And solving for
we got:
