Question

A potter's wheel has the shape of a solid uniform disk of mass 13.0 kg and radius 1.25 m. It spins about an axis perpendicular to the disk at its center. A small 1.7 kg lump of very dense clay is dropped onto the wheel at a distance 0.63 m from the axis. What is the moment of inertia of the system about the axis of spin?

Answer 1

Answer:$10.82 kg-m^2$

Explanation:

Given

Mass of solid uniform disk $M=13 kg$

radius of disk $r=1.25 m$

mass of lump $m=1.7 kg$

distance of lump from axis $r_0=0.63$

Moment of inertia is the distribution of mass from the axis of rotation

Initial moment of inertia of disk $I_1=\frac{Mr^2}{2}$

$I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2$

Final moment of inertia $I_f$=Moment of inertia of disk+moment of inertia of lump about axis

$I_f=\frac{Mr^2}{2}+mr_0^2$

$I_f=10.15+1.7\times 0.63^2$

$I_f=10.15+0.674$

$I_f=10.82 kg-m^2$