Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Figure 10-34a shows a disk that can rotate about an axis at a radial distance h from the center of the disk. Figure 10-34b gives the rotational inertia l of the disk about the axis as a function of that distance h, from the center out to the edge of the disk. The scale on the l axis is set by $l_{A} = 0.050 k g . m^{2}$ and $l_{B} = 0.050 k g . m^{2}$ . What is the mass of the disk?

The mass of the disk is, 2.5kg

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Understanding the given information

Moment of inertia at $h_{A} = 0 m$ is $l_{A} = 0.050 k g . m^{2}$
Moment of inertia at $h_{B} = 0.2 m$ is, $l_{B} = 0.150 k g . m^{2}$ .

Step 2: Concept and Formula used in the given question

By applying the parallel axis theorem for two values of moment of inertia as given in the graph we will get two equations. By solving them we can find mass of the disk.

Parallel axis theorem,

$l = l_{c o m} + m h^{2}$

Step 3: Calculation for the mass of the disk

To calculate mass of the disk we will apply Parallel axis theorem for l_A and l_B

$l_{A} = l_{c o m} + m h_{A}^{2}$ …(1)

$l_{B} = l_{c o m} + m h_{B}^{2}$ …(2)

l_com is the moment of inertia of the disk about its center of mass.

Subtracting equation (2) from (1)

$\begin{array}{rcl} l_{B} - l_{A} & = & m (h_{B}^{2} - h_{A}^{2}) \\ m & = & \frac{l_{B} - l_{A}}{h_{B}^{2} - h_{A}^{2}} \end{array}$

Substitute all the value in the above equation.

We can use the values of h_A,h_B from the given graph

$\begin{array}{rcl} m & = & \frac{0.150 k g . m^{2} - 0.050 k g . m^{2}}{{(0.2 m)}^{2} - {(o m)}^{2}} \\ = & 2.5 k g \end{array}$

Hence the mass of the disk is, 2.5kg.

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Fundamentals Of Physics

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Step by Step Solution

Step 1: Understanding the given information

Step 2: Concept and Formula used in the given question

Step 3: Calculation for the mass of the disk

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