Answer to Question #201458 in Mechanics | Relativity for Michelle Tammi

Question

A disc with a mass of 1.21 kg rotates around an axis z with an initial angular velocity of 3.25 rad/s. On top of the disc (which has an inertia radius of 2.55m) is a rectangular disc with mass 1.21kg, base 3.163m and height 3.206m and rotates at the same angular velocity as the lower disc 3.25 rad/s. The axis of rotation passes through the center of mass of both discs. Calculate the total rotational energy of the discs.

Answer 1

Explanations & Calculations

• What we need to find is the total rotational energy, which is as follows

$\qquad\qquad \begin{aligned} \small E_r&=\small \frac{1}{2}I_{disc}\omega^2+\frac{1}{2}I_{plate}\omega^2\\ &=\small \frac{\omega^2}{2}(I_{d}+I_{p})\cdots(1) \end{aligned}$

• Angular velocity is known & it's about calculating the moments of inertias of both the disc & the plate.

• For the disc, its mass is given & the radius of gyration instead of the actual radius is given. Then we need to calculate the actual radius.
• For a disc

$\qquad\qquad \begin{aligned} \small K&=\small \frac{r_d}{\sqrt2}\\ \small r_d&=\small \sqrt2 \times2.55\,m\\ &=\small 3.61\,m \end{aligned}$

• Then its inertia is

$\qquad\qquad \begin{aligned} \small I_d&=\small \frac{1}{2}mr^2=0.5\times1.21\,kg\times3.61^2m^2\\ &=\small 7.88\,kgm^2 \end{aligned}$

• For the rectangular disc, inertia about the axis along the longer edge is

$\qquad\qquad \begin{aligned} \small I_{center}&=\small \frac{1}{3}m(a^2)\\ &=\small \frac{1}{12}\times1.21\,kg\times(3.163^2m^2)\\ &=\small 1.01\,kgm^2 \end{aligned}$

• As the question describes, it rests on the circular disc vertically up, the z-axis being the central axis. Then the needed moment of inertia about that axis can be calculated with the help of the parallel axis theorem.
• Then it is,

$\qquad\qquad \begin{aligned} \small I_{z}&=\small I+mx^2\\ &=\small 1.01+1.21\times(\frac{3.163m}{2})^2\\ \small I_{plate}&=\small I_z=4.04\,kgm^2 \end{aligned}$

• Finally, the answer can be obtained by the equation 1,

$\qquad\qquad \begin{aligned} \small E_k&=\small \frac{(3.25\,rads^{-1})^2}{2}\times(7.88+4.04)\\ &=\small \bold{62.95\,J} \end{aligned}$