91Ó°ÊÓ

i) The disk $A$has radius$R$

ii) Hub of disk $A$has radius$0.5000R$

iii) The disk $B$has radius$R_B=0.2500R$

iv) The disk$C$has radius$R_c=2.000R$

v) Density of disk $B$and diskis$C$same as$\mathrm{ÒÏ}$

Use the expression of angular momentum in terms of rotational inertia and angular velocity. Find the mass of each disk by using the expression of density. All disks are connected by the same belt at the rim as well as its hub; hence their linear velocity will be the same. We use the expression of relation between linear velocity, angular velocity, and their radius.

Formulae:

$v=R\mathrm{Ó\neg }$

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

$\mathrm{ÒÏ}=\frac{M}{V}$

$L=I\mathrm{Ó\neg }$

The linear speed at the rim of disk$A$ must equal the linear speed at the rim of disk $C$.

$\begin{array}{c}{v}_A=v_C\\ R{\mathrm{Ó\neg }}_A=2.000R{\mathrm{Ó\neg }}_C\end{array}$

${\mathrm{Ó\neg }}_A=2.000{\mathrm{Ó\neg }}_C$

…(¾±)

The linear speed at the hub of diskmust equal the linear speed at the rim of disk

$\begin{array}{c}{v}_A=v_B\\ 0.5000R{\mathrm{Ó\neg }}_A=0.2500R{\mathrm{Ó\neg }}_B\end{array}$

$⇒{\mathrm{Ó\neg }}_A=\frac{1}{2}{\mathrm{Ó\neg }}_B$ …(¾±¾±)

From equations (i) and (ii) as

$2.000{\mathrm{Ó\neg }}_C=\frac{1}{2}{\mathrm{Ó\neg }}_B$

$⇒{\mathrm{Ó\neg }}_B=4{\mathrm{Ó\neg }}_C$

The angular momenta depend on their angular velocities as well as their moment of inertia.

$L=I\mathrm{Ó\neg }$

The moment of inertia of the disk about an axis and passing through its center is

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

If$h$is the thickness and$\mathrm{ÒÏ}$is the density of each disk, then

$\begin{array}{l}\mathrm{ÒÏ}=\frac{M}{\mathrm{Ï€}{R}^{2}h}\\ M=\mathrm{ÒÏ}\mathrm{Ï€}{R}^{2}h\end{array}$

The ratio of angular momenta of disk$C$to the disk$B$is

$\begin{array}{c}\frac{L_C}{L_B}=\frac{I_C{\mathrm{Ó\neg }}_C}{I_B{\mathrm{Ó\neg }}_B}\\ \frac{L_C}{L_B}=\frac{\left(\frac{1}{2}{M}_C{R}_C^2\right){\mathrm{Ó\neg }}_C}{\left(\frac{1}{2}{M}_B{R}_B^2\right){\mathrm{Ó\neg }}_B}\\ \frac{L_C}{L_B}=\frac{\left(\frac{1}{2}\mathrm{ÒÏ}\mathrm{Ï€}{R}_C^{2}h{R}_C^2\right){\mathrm{Ó\neg }}_C}{\left(\frac{1}{2}\mathrm{ÒÏ}\mathrm{Ï€}{R}_B^{2}h{R}_B^2\right)4{\mathrm{Ó\neg }}_C}\\ \frac{L_C}{L_B}=\frac{R_C^4}{4R_B^4}\\ \frac{L_C}{L_B}=\frac{{\left(2.000R\right)}^4}{4{\left(0.2500R\right)}^4}\end{array}$

$⇒\frac{L_C}{L_B}=1024$