Question

In Fig. 10 - 37, two particles, each with mass m = 0.85KG , are fastened to each other, and to a rotation axis at O , by two thin rods, each with length d = 5.6cm and mass M = 1.2kg . The combination rotates around the rotation axis with the angular speed v = 0.30rad/s . Measured about O, what are the combination’s (a) rotational inertia and (b) kinetic energy?

Short answer

1. The combination rotational inertia of the system is, 0.023kg.m² .
2. The combination kinetic energy of the system is, $1.1\times {10}^{-3}J$.

Step-by-step solution

Understanding the given information

1. The mass of each particle is, m = 0.85kg .
2. The rod length is, d = 5.6cm .
3. The mass of rod is, M = 1.2kg .
4. The angular speed of the system is, $\omega =0.30rad/s$.

Concept and Formula used int the given question

By using the concept of moment of inertia we can calculate the moment of inertia at various points. Moment of inertia of the system is the mass times perpendicular distance square at that point. From the moment of inertia of the system we can calculate the kinetic energy of the system and is given as follows.

1. Moment of inertia at point is l = Md²
2. For a rod at one end moment of inertia is $\mathit{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}\mathit{M}{\mathit{d}}^{\mathbf{2}}$
3. Kinetic energy of the system is $\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{l}{\mathit{\omega }}^{\mathbf{2}}$
4. Parallel axis theorem $\mathit{l}\mathbf{=}{\mathit{l}}_{\mathbf{c}\mathbf{o}\mathbf{m}}\mathbf{+}\mathit{M}{\mathit{d}}^{\mathbf{2}}$
   Where d is the perpendicular distance.

(a) Calculation for the rotational inertia

According to parallel axis theorem moment of inertia is

$l=l_{com}+m{d}^2$

For our system we have,

l₁ for first rod, l₂ for mass 1, l₃ for second rod, l₄ for mass 2.

$\begin{array}{rcl}l& =& l_1+l_2+l_3+l_4\\ l_1& =& \left(\frac{1}{12}M{d}^2+M{\left(\frac{1}{2}d\right)}^2\right)\\ l_2& =& m{d}^2\\ l_3& =& \left(\frac{1}{12}M{d}^2+M{\left(\frac{3}{2}d\right)}^2\right)\\ l_4& =& m{\left(2d\right)}^2\\ l& =& \frac{4}{12}M{d}^2+\frac{37}{12}M{d}^2+m{d}^2+4m{d}^2\left(\frac{1}{12}M{d}^2+M{\left(\frac{1}{2}d\right)}^2\right)+m{d}^2\\ & & +\left(\frac{1}{12}M{d}^2+M{\left(\frac{3}{2}d\right)}^2\right)+m{\left(2d\right)}^2\\ & =& \frac{4}{12}M{d}^2+\frac{28}{12}M{d}^2+m{d}^2+4m{d}^2\\ l& =& \frac{8}{3}M{d}^2+5m{d}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}l& =& \frac{8}{3}\times 1.2kg\times {\left(0.056m\right)}^2+5\times 0.85kg\times {\left(0.056m\right)}^2\\ & =& 0.02336kg.m^2\\ & \approx & 0.023kg.m^2\end{array}$

Hence the rotational inertia is, 0.023kg.m² .

(b) Calculation for the kinetic energy

Kineticenergy of the system is

$K=\frac{1}{2}l{\omega }^2$

Where $\omega$is the angular speed of the particle.

Substitute all the value in the above equation.

role="math" localid="1661094304990" $\begin{array}{rcl}K& =& \frac{1}{2}\times 0.023336kg.m^2\times {\left(0.30rad/s\right)}^2\\ & =& 1.05\times {10}^{-3}J\\ & \approx & 1.1\times {10}^{-3}J\end{array}$

Hence the kinetic energy is, $\begin{array}{rcl}& & 1.1\times {10}^{-3}J\end{array}$.