• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q36P

Expert-verified Found in: Page 323 ### Fundamentals Of Physics

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${\mathbf{0}}{\mathbf{.5000}}{\mathbit{R}}$ ; disk B has radius ${\mathbf{0}}{\mathbf{.2500}}{\mathbit{R}}$; and disk C has radius ${\mathbf{2}}{\mathbf{.000}}{\mathbit{R}}{\mathbf{.}}{\mathbf{}}$Disks 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 disk C to that of disk B? The ratio of the magnitude of the angular momentum of disk $C$ to that of disk $B$ is $\frac{{L}_{C}}{{L}_{B}}=1024$

See the step by step solution

## Step 1: Identification of given data

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$\rho$

## Step 2: To understand the concept

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:

$v=R\omega$

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

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

$L=I\omega$

## Step 3: Determining ωAand ωBin terms of ωC

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

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

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

From equations (i) and (ii) as

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

$⇒{\omega }_{B}=4{\omega }_{C}$

## Step 4: Determining the ratio of the magnitude of the angular momentum of disk  Cto that of disk B

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

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

$⇒\frac{{L}_{C}}{{L}_{B}}=1024$

## Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades. 