91Ó°ÊÓ

The length is,$L=1.0000m$.

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

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, $\mathit{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}\mathit{M}{\mathit{L}}^{\mathbf{2}}$

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,

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

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

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

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

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