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Q36P

Expert-verifiedFound in: Page 323

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Figure shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks A and C. Another belt runs around a central hub on disk A and the rim of disk B. The belts move smoothly without slippage on the rims and hub. Disk A has radius R; its hub has radius${\mathbf{0}}{\mathbf{.5000}}{\mathit{R}}$ ; disk B has radius ${\mathbf{0}}{\mathbf{.2500}}{\mathit{R}}$**

The ratio of the magnitude of the angular momentum of disk $C$ to that of disk $B$ is $\frac{{L}_{C}}{{L}_{B}}=1024$

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$\rho $

**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\omega $

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

$\rho =\frac{M}{V}$

$L=I\omega $

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

$\begin{array}{c}{v}_{A}={v}_{C}\\ R{\omega}_{A}=2.000R{\omega}_{C}\end{array}$

${\omega}_{A}=2.000{\omega}_{C}$

…(i)

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

$\begin{array}{c}{v}_{A}={v}_{B}\\ 0.5000R{\omega}_{A}=0.2500R{\omega}_{B}\end{array}$

$\Rightarrow {\omega}_{A}=\frac{1}{2}{\omega}_{B}$ …(ii)

From equations (i) and (ii) as

$2.000{\omega}_{C}=\frac{1}{2}{\omega}_{B}$

$\Rightarrow {\omega}_{B}=4{\omega}_{C}$

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

$L=I\omega $

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$\rho $ is the density of each disk, then

$\begin{array}{l}\rho =\frac{M}{\pi {R}^{2}h}\\ M=\rho \pi {R}^{2}h\end{array}$

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

$\begin{array}{c}\frac{{L}_{C}}{{L}_{B}}=\frac{{I}_{C}{\omega}_{C}}{{I}_{B}{\omega}_{B}}\\ \frac{{L}_{C}}{{L}_{B}}=\frac{\left(\frac{1}{2}{M}_{C}{R}_{C}^{2}\right){\omega}_{C}}{\left(\frac{1}{2}{M}_{B}{R}_{B}^{2}\right){\omega}_{B}}\\ \frac{{L}_{C}}{{L}_{B}}=\frac{\left(\frac{1}{2}\rho \pi {R}_{C}^{2}h{R}_{C}^{2}\right){\omega}_{C}}{\left(\frac{1}{2}\rho \pi {R}_{B}^{2}h{R}_{B}^{2}\right)4{\omega}_{C}}\\ \frac{{L}_{C}}{{L}_{B}}=\frac{{R}_{C}^{4}}{4{R}_{B}^{4}}\\ \frac{{L}_{C}}{{L}_{B}}=\frac{{\left(2.000R\right)}^{4}}{4{\left(0.2500R\right)}^{4}}\end{array}$

$\Rightarrow \frac{{L}_{C}}{{L}_{B}}=1024$

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