What is speed of yo-yo?

Question

The string in a yo-yo is wound around an axle of radius 0.545 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.274 kg and outer radius 1.96 cm. Starting from rest, it rotates and falls a distance of 1.50 m (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle.

(a) What is the speed of the yo-yo when it reaches the distance of 1.50 m?
_______m/s
(b) How long does it take to fall? [Hint: The translational and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]
______s

Answer 1

a) Let's use conservation of energy
m*g*h=.5*I*w^2+.5*m*v^2
since the yo-yo is spun by the torque on
the axle,
w=v/r, where r is the radius of the axle and
I=m*R^2/2, where R is the outer radius

so
m*g*h=.5*.*R^2*v^2/(2*r)+.5*m*v^2
v=sqrt((2*g*h)/(R^2/(2*r)+1))
v=5.33 m/s

b)T*r=I*alpha
where T is the tension in the string
and
m*g-T=m*a
again, the connection of the axle relates
alpha=a/r
where r is the radius of the axle

so
T=m*R^2*a/(2*r^2)
T=m*(g-a)
a=g/(R^2/(2*r^2)+1)
a=1.31 m/s^2

using
y(t)=-.5*a*t^2
when y(t)=-1.5
t=1.51 seconds

j