Question

Two thin disks are attached as shown to the right. The angle θ = 55.7°, and the masses and radii of the disks are listed in the table below. Disk radius (cm) Mass (Kg) A 13.8 B 8.70 0.566 0.245 What is the moment of inertia of this composite object about an axis perpendicular to the screen, through point p? Number 0.2557 kg m

Answer 1

r3 r1

r1=0.138m, r2=0.138+0.0870 = 0.225m

Let us find r3 now,

Use cosine law in the triangle in the figure above,

Φ=180-55.7=124.3 deg

r3^2=r1^2+r2^2-2*r1*r2*cosΦ

r3^2=0.138^2*0.225^2-2*0.138*0.225*cos124.3

r3=0.1896m

Moment of inertia of a disk = 1/2mr^2

Moment of inertia of disk A about its center = I_A=1/2*m_A*r1^2 =1/2*0.566*0.138^2 = 0.005389 kg*m^2

By parallel axis theorem,

Moment of inertia of disk A about point p = I_AP= I_A + m_A*r1’^2 = 0.005389 + 0.566*0.138^2 = 0.01617 kg*m^2

Moment of inertia of disk B about its center = I_B=1/2*m_B*r2’^2 =1/2*0.245*0.0.0870^2 = 0.0009272

By parallel axis theorem,

Moment of inertia of disk A about point p = I_BP= I_B + m_B*r3’^2 = 0.0009272+ 0.245*0.225^2 =

0.01333kg*m^2

Moment of inertia of both disks about point p = I_AP+ I_BP= 0.01617+ 0.01333 = 0.0295 kg*m^2