Question

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.

Short answer

a) Rotational kinetic energy along the axis passing through point P and perpendicular to plane is $3.3\text{\hspace{0.17em}J}$.

b) Rotational kinetic energy about axis passing through P and that lies in plane of figure is 2.9 J.

Step-by-step solution

Step 1: Given

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

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

iii) Angle $\theta =30°$

Determining the concept

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.

(a) Determining the rotational kinetic energy about the axis passing through p and perpendicular to the plane

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 .

(b) Determining the rotational kinetic energy about the axis passing through p and lies in the plane of figure

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$2.9\text{\hspace{0.17em}J}$.