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Q36P

Expert-verifiedFound in: Page 288

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Figure 10-34 a**

The mass of the disk is, 2.5kg

- Moment of inertia at ${h}_{A}=0m$ is ${l}_{A}=0.050kg.{m}^{2}$
- Moment of inertia at ${h}_{B}=0.2m$ is, ${l}_{B}=0.150kg.{m}^{2}$ .

**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,**

** ${\mathit{l}}{\mathbf{=}}{{\mathit{l}}}_{\mathbf{c}\mathbf{o}\mathbf{m}}{\mathbf{+}}{\mathit{m}}{{\mathit{h}}}^{{\mathbf{2}}}$**

To calculate mass of the disk we will apply Parallel axis theorem for l_{A} and l_{B}

${l}_{A}={l}_{com}+m{h}_{A}^{2}$ …(1)

${l}_{B}={l}_{com}+m{h}_{B}^{2}$ …(2)

l_{com} is the moment of inertia of the disk about its center of mass.

Subtracting equation (2) from (1)

$\begin{array}{rcl}{l}_{B}-{l}_{A}& =& m\left({h}_{B}^{2}-{h}_{A}^{2}\right)\\ m& =& \frac{{l}_{B}-{l}_{A}}{{h}_{B}^{2}-{h}_{A}^{2}}\end{array}$

Substitute all the value in the above equation.

We can use the values of h_{A},h_{B} from the given graph

$\begin{array}{rcl}m& =& \frac{0.150kg.{m}^{2}-0.050kg.{m}^{2}}{{\left(0.2m\right)}^{2}-{\left(om\right)}^{2}}\\ & =& 2.5kg\end{array}$

Hence the mass of the disk is, 2.5kg.

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