Question

The figure shows three rotating, uniform discs that are coupled by belts. One belt runs around the rims of discs $A$ and $C$. Another belt runs around a central hub on the disc $A$ and the rim of the disc $B$. The belts move smoothly without slippage on the rims and hub. Disc $A$ has radius $R$; its hub has radius $0.500R$; disc $B$ has radius $0.250R$ and disc $C$ has radius $2.00R$. Discs $B$ and $C$ have the same density $($mass per unit volume$)$ and thickness. What is the ratio of the magnitude of the angular momentum of the disc $C$ to that of the disc $B$ ?

• A. $1224:1$
• B. $1024:1$
• C. $1424:1$
• D. $1624:1$

Answer 1

The correct option is B $1024:1$

Given:
Radius of disc $A$, $r_A=R$
Radius of hub of disc $A$, $r_{\text{Ah}}=0.500R$
Radius of disc $B$, $r_B=0.250R$
Radius of disc $C$, $r_C=2.00R$

The motion of the all three discs can be related by linear velocity of the belts.

Let the angular velocities of the discs $A,B$ and $C$ be ${\omega }_A$, ${\omega }_B$ and ${\omega }_C$ respectively.

Then, for the hub of disc $A$ and disc $B$,
$\Rightarrow {\omega }_A{r}_{\text{Ah}}={\omega }_B{r}_B$
$\Rightarrow {\omega }_A\times 0.500R={\omega }_B\times 0.250R$
$\Rightarrow 2{\omega }_A={\omega }_B\dots \dots \left(1\right)$

Now for disc $A$ and $C$,
${\omega }_A{r}_A={\omega }_C{r}_C$
$\Rightarrow {\omega }_{A}R={\omega }_C\times 2.00R$
$\Rightarrow {\omega }_A=2{\omega }_C\dots \dots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$\Rightarrow {\omega }_B=4{\omega }_C\dots \dots \left(3\right)$

Let the thickness and density $[$of each disc $B$ and $C$$]$ be $t$ and $\rho$ respectively.

Angular momentum of a disc is given by $L=I\omega =\frac{m{r}^2}{2}\omega$

Now, $\frac{\text{Angular momentum of disk}C}{\text{Angular momentum of disk}B}=\frac{0.5\left(\rho \pi r_C^{2}t\right)\times r_C^2{\omega }_C}{0.5\left(\rho \pi r_B^{2}t\right)\times r_B^2{\omega }_B}$
$\Rightarrow \frac{L_C}{L_B}=\frac{r_C^4{\omega }_C}{r_B^4{\omega }_B}$
$\Rightarrow \frac{L_C}{L_B}=\frac{(2R{)}^4{\omega }_C}{(0.25R{)}^4\times 4{\omega }_C}$ $[$from $\left(3\right)]$
$\Rightarrow \frac{L_C}{L_B}=1024:1$