A yo-yo is spun from rest by pulling on the string with a constant tension of 2.0 N. The radius of the inner rod on which the string is strung around is 0.50 cm. The tension is applied for 5.0 seconds after which the yo-yo is observed to spin with an angular velocity of 15 rad/sec.

a.) From the information above, what is the moment of inetia for the yo-yo?

b.)What is the total angle the yo-yo has traveled through in these 5.0 seconds?

Now you press your finger against the outer rim of the yo-yo (which has a radius of 4.0 cm) to bring it to a stop. You apply a constant force of 2.0 N directed perpendicular to the rim of the yo-yo. The coefficient of kinetic friction between your finger and the edge of the yo-yo is 0.80.

c.)How long does it take for the yo-yo to come to a stop?

I got parts a and b but am struggling with c

Moments about center of yoyo shaft

torque = 2*(.5/100) = .01 NM

omega = alpha * t
but
alpha = Torque/ I
so
omega =Torque * t/I
or
I = Torque * t / omega

= .01 * 5 / 15
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angle = (1/2) alpha t^2
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now for part c I assume you no longer apply that 2 N up at a radius of the shaft out because the yoyo is at the bottom and spinning freely in a loop at the end of the string (the problem is not clear on that point)
So
the only torque is friction

torque = 2 Newtons * .8 * .04

alpha = torque/ I

omega = 15 - alpha* t
at stop then
alpha * t = 15
t = 15/alpha