The Yo-Yo Fall

Find the acceleration of a fixed Yo-Yo and a Yo-Yo in free fall (see the images). Moment of inertia is $I=MR^2/2$. What are the fall times and velocities for the two cases after 1 meter of fall. Numerical values are $R=20 mm$ and $r=5 mm$

Compare Yo-Yo Fixed and in Free Fall

For Yo-Yo in free fall

$Mg=T+Ma_1$

string is around small cylinder ($r$) so $T*r=Iepsilon$ with $epsilon=a_1/r$

$Mg=Iepsilon/r +Ma_1$   or $Mg=Ia/r^2 +Ma_1$ so that

$a_1=frac{Mg}{M+(I/r^2)}$

Since $I=MR^2/2$ then

$a_1=frac{g}{1+R^2/2r^2)}$  or $a_1=frac{9.81}{1+(400/(2*25))}=1.09 m/s^2$

For mass M in free fall and yo-yo fixed

$Mg=T+Ma_2$

String is around big cylinder (R) so that $TR=Iepsilon$ with $epsilon=a_2/R$

$Mg=Iepsilon/R+Ma_2$ or $Mg=Ia_2/R^2 +Ma_2$ so that

$a_2=Mg/[M+(I/R^2)]$

since $I=MR^2/2$ then $a_2=g/[1+(1/2)]=(2/3)g=6.54m/s^2$

Fall times are

$t=sqrt {2h/a}$ so that with $h=1 m$ one has $t_1=1.355 s$ and $t_2=0.553 s$

Velocities after 1 m of fall are

$v=sqrt{2ah}$ so that with $h=1$ one has $v_1=1.476 m/s$ and $v_2=3.617 m/s$

Angular accelerations are

$epsilon_1=a_1/r=1.09/0.005=218 (rad/s^2)$

$epsilon_2=a_2/R=6.54/0.02=327 (rad/s^2)$