Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $0 .5000 R$ ; disk B has radius $0 .2500 R$ ; and disk C has radius $2 .000 R .$ 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 $\frac{L_{C}}{L_{B}} = 1024$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of given data

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

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

iii) The disk $B$ has radius $R_{B} = 0.2500 R$

iv) The disk $C$ has radius $R_{c} = 2.000 R$

v) Density of disk $B$ and diskis $C$ same as $ρ$

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:

$v = R ω$

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

$ρ = \frac{M}{V}$

$L = I ω$

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 $C$ .

$\begin{matrix} v_{A} = v_{C} \\ R ω_{A} = 2.000 R ω_{C} \end{matrix}$

$ω_{A} = 2.000 ω_{C}$

…(i)

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

$\begin{matrix} v_{A} = v_{B} \\ 0.5000 R ω_{A} = 0.2500 R ω_{B} \end{matrix}$

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

From equations (i) and (ii) as

$2.000 ω_{C} = \frac{1}{2} ω_{B}$

$\Rightarrow ω_{B} = 4 ω_{C}$

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.

$L = I ω$

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 $ρ$ is the density of each disk, then

$\begin{array}{l} ρ = \frac{M}{π R^{2} h} \\ M = ρ π R^{2} h \end{array}$

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

$\begin{matrix} \frac{L_{C}}{L_{B}} = \frac{I_{C} ω_{C}}{I_{B} ω_{B}} \\ \frac{L_{C}}{L_{B}} = \frac{(\frac{1}{2} M_{C} R_{C}^{2}) ω_{C}}{(\frac{1}{2} M_{B} R_{B}^{2}) ω_{B}} \\ \frac{L_{C}}{L_{B}} = \frac{(\frac{1}{2} ρ π R_{C}^{2} h R_{C}^{2}) ω_{C}}{(\frac{1}{2} ρ π R_{B}^{2} h R_{B}^{2}) 4 ω_{C}} \\ \frac{L_{C}}{L_{B}} = \frac{R_{C}^{4}}{4 R_{B}^{4}} \\ \frac{L_{C}}{L_{B}} = \frac{{(2.000 R)}^{4}}{4 {(0.2500 R)}^{4}} \end{matrix}$

$\Rightarrow \frac{L_{C}}{L_{B}} = 1024$

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Identification of given data

Step 2: To understand the concept

Step 3: Determining ωAand ωBin terms of ωC

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

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Fundamentals Of Physics

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Step by Step Solution

Step 1: Identification of given data

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

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