Four particles, each of mass, $\text{0.20\hspace{0.17em}kg}$, are placed at the vertices of a square with sides of length$\mathbf{\text{0.50m}}$. The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal (Fig$\mathbf{\text{10-64}}$.). (a) What is the rotational inertia of the body about axis A? (b) What is the angular speed of the body about axis A when rod AB swings through the vertical position?

i) Mass of particle is$0.20\text{\hspace{0.17em}kg}$

ii) Length of square is$0.50\text{\hspace{0.17em}m}$

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-

$\mathbf{I}\mathbf{=}{\mathbf{\Sigma mr}}^2$

The rotational kinetic energy is given as-

$\mathbf{KE}\mathbf{=}\frac{1}{2}{\mathbf{I\omega }}^2$

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

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

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

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

Now, rotational inertia is as follows:

$\begin{array}{c}I=m{\left(\text{AB}\right)}^2+m{\left(\text{AD}\right)}^2+m{\left(\text{AC}\right)}^2\\ =0.20\text{\hspace{0.17em}kg}{\left(0.5\text{\hspace{0.17em}m}\right)}^2+0.20\text{\hspace{0.17em}kg}{\left(0.50\text{\hspace{0.17em}m}\right)}^2+0.20\text{\hspace{0.17em}kg}{\left(0.71\text{\hspace{0.17em}m}\right)}^2\\ =0.20\text{kg}.{\text{m}}^2\end{array}$

Hence,rotational inertia of body about axis A is .$0.20\text{\hspace{0.17em}kg}.{\text{m}}^2$

Rotational kinetic energy can be expressed as,

$KE=\frac{1}{2}I{\omega }^2$

And gravitational potential energy of system of four masses is,

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

Now, according to conservation of energy,

$KE=PE$

$\frac{1}{2}I{\omega }^2=\left(4m\right)gL$

localid="1661002600155" $0.5\times 0.20{\text{\hspace{0.17em}kgm}}^{\text{2}}\times {\omega }^2=(4\times 0.20\text{\hspace{0.17em}kg})\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 0.5\text{\hspace{0.17em}\hspace{0.17em}m}$

$\omega =6.3\text{\hspace{0.17em}rad/s}$

Hence,angular speed of body about axis A is.$6.3\text{\hspace{0.17em}}\text{rad/s}$