91Ó°ÊÓ

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

ii) Length$L=0.6\text{″¾}$

iii) Angle $\mathrm{θ}=30°$

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-

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

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

Total moment of inertia is as follows:

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

$I=\left(1.6\text{ k²µ}\right){\left(1.2\text{″¾}\right)}^2+\left(3.2\text{ k²µ}\right){\left(0.6\text{″¾}\right)}^2+\left(3.2\text{ k²µ}\right){\left(0.6\text{″¾}\right)}^2$

$I=4.608\text{ k²µ}.{\text{m}}^2$

Now, rotational kinetic energy is as follows:

$K{E}_{rotational}=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

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

$K{E}_{rotational}=3.3\text{ J}$

Hence,rotational kinetic energy about axis passing through $3.3\text{ J}$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:

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

$I=\left(3.2\text{ k²µ}\right){\left(0.52\text{″¾}\right)}^2+\left(3.2\text{ k²µ}\right){\left(0.52\text{″¾}\right)}^2+\left(1.6\text{ k²µ}\right){\left(1.2\text{″¾}\right)}^2$

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

Now, rotational kinetic energy is follow:

$K{E}_{rotational}=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

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

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

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