Question

A child of mass m = 40 kg is sitting on a merry-go-round (disk) of mass M = 120 kg and radius R = 1.0 m at a distance r=0.50 m from the axis of rotation. Calculate the moment of inertia of this system, in kg m2. I disk IMR2. Do not type in any units. m M

Answer 1

We have,
Mass of the child $\rightarrow m=40\ kg$
Mass of the disk $\rightarrow M=120\ kg$
Radius of disk $\rightarrow R=1 \ m$
Radius at which the child is sitting
►r= 0.50 m

The total moment of inertia of the system is the sum of moment of inertia of disk $I_{disk}$ and moment of inertia of the child $I_{child}$ .

The moment of inertia of the disk $I_{disk}$ is calculated as:

$I_{disk}=\frac{1}{2}MR^{2}$

$\therefore I_{disk}=\frac{1}{2}\times120\times1^{2}$

${\color{Blue} \therefore I_{disk}=60\ kg\ m^{2}}$

The moment of inertia of the child $I_{child}$ is calculated by considering the child to be equivalent to a point mass from the same axis of rotation of the disk:

$I_{child}=mr^{2}$

$\therefore I_{child}=40\times0.50^{2}$

${\color{Blue} \therefore I_{child}=10\ kg\ m^{2}}$

Therefore, total moment of inertia of the system is calculated as:

$I=I_{disk}+I_{child}$

$\therefore I=60+10$

${\color{Blue} \therefore I=70\ kg\ m^{2}}$

Total moment of inertia of the system is ${\color{Blue} I=70\ kg\ m^{2}}$ .