Answer
$T_1=1.20N;T_2=0.2153N;m=21.9g$
Work Step by Step
We can find $T_1, T_2$ and $m$ as follows:
The force acting on the yo-yo as
$T_1-T_2=Mg$.....eq(1)
The net torque is given as
$T_1r-T_2R=0$
$T_1r=T_2R$
$T_1r=T_2(5.60r)$
$T_1=5.60T_2$......eq(2)
We plug in this value in eq(1) to obtain:
$5.60T_2-T_2=Mg$
$T_2(5.60-1)=Mg$
This simplifies to:
$T_2=\frac{Mg}{4.60}$
$T_2=\frac{(0.101Kg)(9.81m/s^2)}{4.60}$
$T_2=0.2153N$
From eq(1)
$T_1=5.060(0.2153N)$
$T_1=1.20N$
Now the mass can be determined as
$m=\frac{T_2}{g}$
We plug in the known values to obtain:
$m=\frac{0.2153N}{9.8m/s^2}$
$m=0.0219Kg$
$m=21.9g$
