Question

Figure 10-34ashows a disk that can rotate about an axis at a radial distance hfrom the center of the disk. Figure 10-34b gives the rotational inertia lof 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 laxis is set by ${\mathit{l}}_{\mathbf{A}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{050}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$and${\mathit{l}}_{\mathbf{B}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{050}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$. What is the mass of the disk?

Short answer

The mass of the disk is, 2.5kg

Step-by-step solution

Understanding the given information

1. Moment of inertia at $h_A=0m$is $l_A=0.050kg.m^2$
2. Moment of inertia at $h_B=0.2m$is, $l_B=0.150kg.m^2$.

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,

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

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

$l_A=l_{com}+m{h}_A^2$ …(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.