Question

Figure 10-36shows an arrangement of 15identical disks that have been glued together in a rod-like shape of length L = 1.0000M and (total) massM = 100.0mg. The disks are uniform, and the disk arrangement can rotate about a perpendicular axis through its central disk at point O . (a) What is the rotational inertia of the arrangement about that axis? (b) If we approximated the arrangement as being a uniform rod of mass Mand length L , what percentage error would we make in using the formula in Table 10-2eto calculate the rotational inertia?

Short answer

1. The rotational inertia of the arrangement about the axis is$8.3518\times {10}^{-6}kg.m^2$.
2. The percentage error made in the formula for M.I of uniform rod about its COM is 0.22

Step-by-step solution

Understanding the given information

The length is,$L=1.0000m$.

The mass is, $M=100.0mg=0.0001kg$.

Concept and Formula used in the given question

You can find the torque acting on the pulley from the applied force using the corresponding formula. Then using the relation between angular acceleration and the torque we can find the angular acceleration. Then integrating it with respect to t we can find the angular speed of the pulley at t = 3s.

M.I of the disk about the axis passing through its COM is, ${\mathit{l}}_{\mathbf{c}\mathbf{o}\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{R}}^{\mathbf{2}}$

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

M.I of an uniform rod about its COM is, $\mathit{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}\mathit{M}{\mathit{L}}^{\mathbf{2}}$

(a) Calculation for the rotational inertia of the arrangement about that axis

Let mass and radius of each disk be m and R respectively.

M.I of the disk about the axis passing through its COM is

$l_{com}=\frac{1}{2}m{R}^2$

According to the parallel axis theorem,

$l=l_{com}+M{R}^2$

Applying to the given system,

The M.I of 15 disks about point O is,

$\begin{array}{rcl}l& =& 15\left(\frac{1}{2}m{R}^2\right)+2\left(m{\left(2R\right)}^2\right)+2\left(m{\left(4R\right)}^2\right)+2\left(m{\left(6R\right)}^2\right)+2\left(m{\left(8R\right)}^2\right)+2\left(m{\left(10R\right)}^2\right)+\\ & & 2\left(m{\left(12R\right)}^2\right)+2\left(m{\left(14R\right)}^2\right)\\ & =& 1127.5m{R}^2\end{array}$

But,

$$\begin{gathered} m=\frac{M}{15} \\ R=\frac{L}{30} \end{gathered}$$

Hence,

$\begin{array}{rcl}l& =& {1127.5\left(\frac{M}{15}\right)\left(\frac{L}{30}\right)}^2\\ & =& 1127.5\left(\frac{0.0001kg}{15}\right){\left(\frac{1.00m}{30}\right)}^2\\ & =& 8.3518\times {10}^{-6}kg.m^2\end{array}$

Therefore, Rotational inertia (M.I) of the arrangement about the axis is 8.3518x10^-6kg.m².

(b) Calculation for the percentage error to calculate the rotational inertia

M.I of an uniform rod about its COM is,

$\begin{array}{rcl}l& =& \frac{1}{12}M{L}^2\\ & =& 8.3333\times {10}^{-6}kg.m^2\\ Percentageerror& =& \frac{\left(8.3518\times {10}^{-6}kg.m^2-8.3333\times {10}^{-6}kg.m^2\right)}{8.3518\times {10}^{-6}kg.m^2}\times 100\\ & =& 0.22\%\end{array}$

Therefore, Percentage error made in the formula for M.I of uniform rod about its COM is 0.22