The rigid object shown in Fig.$\mathbf{\text{10-62}}$consists of three ballsand three connecting rods, with$\mathbf{M}\mathbf{\text{=1.6kg,}}\mathbf{L}\mathbf{\text{=0.60m}}$and$\mathbf{\theta }\mathbf{=}\mathbf{\text{30°}}$. The balls may be treated as particles, and the connecting rods have negligible mass. Determine the rotational kinetic energy of the object if it has an angular speed of$\mathbf{\text{1.2\hspace{0.17em}rad/s}}$about (a) an axis that passes through point P and is perpendicular to the plane of the figure and (b) an axis that passes through point P, is perpendicular to the rod of length$\mathbf{\text{2L}}$, and lies in the plane of the figure.

i) Mass of object$M=1.6\text{kg}$,

ii) Length$L=0.6\text{\hspace{0.17em}m}$

iii) Angle $\theta =30°$

Use the formula for rotational kinetic energy in terms of inertia and angular velocity to find the rotational kinetic energy about different axes.

The formula for kinetic energy of a rotating body is expressed as-

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

where, KE is kinetic energy, Iis moment of inertia and$\mathbf{\omega }$ is angular velocity.

Total moment of inertia is as follows:

$I=M{\left(2L\right)}^2+2M{\left(L\right)}^2+2M{\left(L\right)}^2$

$I=\left(1.6\text{\hspace{0.17em}kg}\right){\left(1.2\text{\hspace{0.17em}m}\right)}^2+\left(3.2\text{\hspace{0.17em}kg}\right){\left(0.6\text{\hspace{0.17em}m}\right)}^2+\left(3.2\text{\hspace{0.17em}kg}\right){\left(0.6\text{\hspace{0.17em}m}\right)}^2$

$I=4.608\text{\hspace{0.17em}kg}.{\text{m}}^2$

Now, rotational kinetic energy is as follows:

$K{E}_{rotational}=\frac{1}{2}I{\omega }^2$

$K{E}_{rotational}=\frac{1}{2}\times {4.608}^2\times {1.2}^2$

$K{E}_{rotational}=3.3\text{\hspace{0.17em}J}$

Hence,rotational kinetic energy about axis passing through $3.3\text{\hspace{0.17em}J}$P and perpendicular to plane is .

Rotational kinetic energy about the axis passing through P and lies in the planeof the figure.

Distance of masses 2Mfrom the axisof rotation is x and y,

$x=y=L\cos 30$

$x=y=0.6\cos 30°$

$x=y=0.52\text{m}$

Now, moment of inertia is as follow:

$I=2M{\left(x\right)}^2+2M{\left(Y\right)}^2+M{\left(2L\right)}^2$

$I=\left(3.2\text{\hspace{0.17em}kg}\right){\left(0.52\text{\hspace{0.17em}m}\right)}^2+\left(3.2\text{\hspace{0.17em}kg}\right){\left(0.52\text{\hspace{0.17em}m}\right)}^2+\left(1.6\text{\hspace{0.17em}kg}\right){\left(1.2\text{\hspace{0.17em}m}\right)}^2$

$I=4.034\text{kg}.{\text{m}}^2$

Now, rotational kinetic energy is follow:

$K{E}_{rotational}=\frac{1}{2}I{\omega }^2$

$K{E}_{rotational}=\frac{1}{2}\times 4.034\times {1.2}^2$

$K{E}_{rotational}=2.9\text{J}$

Hence, rotational kinetic energy about axis passing through P and lies in plane of figure is$2.9\text{\hspace{0.17em}J}$.