A uniform disk of radius 0.529 m and unknown mass is constrained to rotate about a perpendicular axis through its center...

Radius of the disk $\left(R\right)=0.529 \text{m}$

Tangential force applied on the rim $\left(F\right)=0.237 \text{N}$

Angular acceleration of the disk and ring system $\left(\alpha \right)=0.119 {\text{rad/s}}^2$

Since the ring is attached to the rim of the disk, the radius of ring will be equal to the radius of the disk.

Therefore, we can find the moment of inertia of this system using the equation;

$RF=I\alpha$

Therefore;

$I=\frac{RF}{\alpha }$

Substituting the values;

$I=\frac{0.529\times 0.237}{0.119}$

$\therefore I=1.053 {\text{kg m}}^2$

The moment of inertia of the disk and ring system is given by;

$I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Hence, we can write for the mass of the disk and the rim as;

$M=\frac{2I}{3R^2}$

Substituting the values;

$M=\frac{2\times 1.053}{3\times {(0.529)}^2}$

$\therefore M=2.51 \text{kg}$

Therefore, the mass of the disk is 2.51 kg.