a. A disk of mass M and radius R has a hole of radius r centered on the axis. Calculate the moment of inertia of the disk.
b. Confirm that your answer agrees with Table 12.2 when r = 0 and when r = R.
c. A 4.0-cm-diameter disk with a 3.0-cm-diameter hole rolls down a 50-cm-long, 20^o ramp. What is its speed at the bottom? What percent is this of the speed of a particle
sliding down a frictionless ramp?

Mass of disk = M

Radius = R

radius of hole = r

Lets assume a small circular strip on the disk with width dx and mass dm.

Lets draw the figure as below.

Mass per unit area of the disk is

$\sigma =\frac{M}{\pi (R^2-r^2)}..............................\left(1\right)$

So, dm can be calculated as

$dm=\sigma (2\pi x)dx$

We can find moment of Inertia as

$dI=dm\times x^2=2\pi \sigma x^{3}dx$

Now integrate and substitute the value of Sigma, we get moment of inertia.

$$\begin{gathered} I=\int dI \\ =2\pi \sigma {\int }_r^R{x}^{3}dx \\ =2\pi \sigma \frac{R^4-r^4}{4} \\ =2\pi \frac{\pi \left(R^2-r^2\right)}{}\frac{\left(R^2+r^2\right)\left(R^2-r^2\right)}{4} \\ =\frac{m\left(R^2+r^2\right)}{2}.....................\left(2\right) \end{gathered}$$

given r=0

and r=R

Substitute r=R in the equation (2) , we get

$\frac{m\left(R^2+r^2\right)}{2}=\frac{m\left(R^2+R^2\right)}{2}=m{R}^2$

This confirms with table 12.2

Diameter of disk= 4.0-cm=0.04m

Diameter of hole = 3.0-cm=0.03m

Length of ramp= 50-cm=0.5 m

Inclination of ramp = 20^o .

First find the moment of inertia of the given disk

$$\begin{gathered} I=\frac{Mkg}{2}\left({(0.04m)}^2+{(0.03m)}^2\right) \\ I=M\times 12.5\times {10}^{-4}kg.m^2 \end{gathered}$$

Vertical displacement of the ramp is $(0.5m)\sin 20°=0.171m$

Angular velocity of the rolling disc is

$\omega =\frac{v}{R}=\frac{v}{4\times {10}^{-2}}$

From the law of energy conservation,

Los sin PE = Gain in KE

$$\begin{gathered} Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{w}^2 \\ Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}M\times (12.5\times {10}^{-4}kg.m^2)\times \left(\frac{{\mathrm{v}}^2}{(16\times {10}^{-4}{\mathrm{m}}^2)}\right) \\ \mathrm{cancel}\mathrm{M}\mathrm{from}\mathrm{both}\mathrm{side} \\ (9.8m/s^2)\times (0.17m)=v^2\left(\frac{1}{2}+\frac{12.5}{32}\right) \\ v=1.37\mathrm{m}/\mathrm{s} \end{gathered}$$

If ramp is frictionless then it will slide and it will not roll.

$$\begin{gathered} Mgh=\frac{1}{2}M{v}^2 \\ v=\sqrt{2\times (9.8m/s^2)\times (17.1\times {10}^{-2}m)}=1.83\mathrm{m}/\mathrm{s} \end{gathered}$$

Now calculate percentage

$\frac{1.37m/sec}{1.83m/sec}\times 100=74.86\%$