91Ó°ÊÓ

i) Mass of particle is$0.20\text{ k²µ}$

ii) Length of square is$0.50\text{″¾}$

The rotational inertia of a body is the quantity that determines how much torque will be needed to produce a particular value of angular acceleration. The formula to find the rotational inertia and rotational kinetic energy are-

The moment of inertia is given as-

$\mathbf{I}\mathbf{=}{\mathbf{Î\pounds ³¾°ù}}^2$

The rotational kinetic energy is given as-

$\mathbf{KE}\mathbf{=}\frac{1}{2}{\mathbf{\pm õÓ\neg }}^2$

where, KE is kinetic energy, I is moment of inertia, m is mass, r is radius, I is moment of inertia and is$\mathbf{Ó\neg }$ angular frequency.

Distance,$\text{AD}=\text{AB}=0.50\text{ }\text{m}$

Distance,$\text{AC}=\sqrt{{0.50}^2+{0.50}^2}$

$\text{AC}=0.71\text{m}$

Now, rotational inertia is as follows:

$\begin{array}{c}I=m{\left(\text{AB}\right)}^2+m{\left(\text{AD}\right)}^2+m{\left(\text{AC}\right)}^2\\ =0.20\text{ k²µ}{\left(0.5\text{″¾}\right)}^2+0.20\text{ k²µ}{\left(0.50\text{″¾}\right)}^2+0.20\text{ k²µ}{\left(0.71\text{″¾}\right)}^2\\ =0.20\text{ }\text{kg}.{\text{m}}^2\end{array}$

Hence,rotational inertia of body about axis A is .$0.20\text{ k²µ}.{\text{m}}^2$

Rotational kinetic energy can be expressed as,

$KE=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

And gravitational potential energy of system of four masses is,

$PE=\left(4m\right)gL$

Now, according to conservation of energy,

$KE=PE$

$\frac{1}{2}I{\mathrm{Ó\neg }}^2=\left(4\text{ }m\right)gL$

localid="1661002600155" $0.5×0.20{\text{ k²µm}}^{\text{2}}×{\mathrm{Ó\neg }}^2=(4×0.20\text{ k²µ})×9.8\text{″¾}/{\text{s}}^{\text{2}}×0.5\text{ ″¾}$

$\mathrm{Ó\neg }=6.3\text{ r²¹»å/²õ}$

Hence,angular speed of body about axis A is.$6.3\text{​â¶Ä‰}\text{rad/s}$