In your job as a mechanical engineer you are designing a flywheel and clutch-plate system like the one in Example 10.11. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed v0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2400 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

The maximum initial kinetic energy of A is given by $k_{0/A/max}=\frac{1}{2}{I}_A{w}_0^2{/}_{A/max}$. By conservation of energy $U_0+K_0+W_{others}=U_1+K_1$.

The flywheel and clutch are rotating so there is no change in the gravitational potential energy implies $U_0=U_1=0and{K}_0=K_1-W_{others}......\left(1\right)$

Find $K_0$as follows:

$\begin{array}{r}{K}_0=K_{0/A}+K_{0/B}\\ =\frac{1}{2}/w_{0/A}^2+\frac{1}{2}/w_{0/B}^2\\ =\frac{1}{2}{l}_{0/A}^2+\frac{1}{2}{I}_B(0{)}^2\\ =\frac{1}{2}{l}_{0/A}^2\end{array}$

Similarly, find $K_1$as follows:

$\begin{array}{r}{K}_1=K_{1/A}+K_{1/B}\\ =\frac{1}{2}{I}_A{w}^2+\frac{1}{2}{I}_B{w}^2\\ =\frac{1}{2}\left(\frac{1}{3}{I}_B\right)+\frac{1}{2}{I}_B{w}^2\\ =\frac{2}{3}{I}_B{w}^2\end{array}$

It is given that $W_{others}=-2400J$then substitute all the value in (1) then,

$\frac{1}{2}I{w^2}_{0/A}=\frac{2}{3}{I}_B{w}^2+2400J$which implies$K_{0/A/max}=\frac{2}{3}{I}_{B{w}^2}+2400J........\left(2\right)$ …… (2)

By the angular momentum conservation, $L_0=L_1$then here, $I_0{w}_0=I_i{w}_1$. Now,

$\begin{array}{r}{I}_A{W}_{0/A}+I_B{W}_{0/B}=I_{A}W+I_{B}W\\ =\left(I_A+I_B\right)W\end{array}$

Since, $I_A=\frac{1}{3}{I}_{B}then,$

$\begin{array}{r}\left(\frac{1}{3}{I}_B\right)W_{0/A}+I_B\left(0\right)=\left(\frac{1}{3}{I}_B\right)w+I_{B}w\\ =\frac{4}{3}{I}_{B}w\end{array}$

From the above equation $w=\frac{1}{4}{w}_{0/A}$. Substitute w in (2) and simplify.

$\begin{array}{r}{K}_{0/A/max}=\frac{2}{3}{I}_B{\left(\frac{1}{4}{w}_{0/A}\right)}^2+2400\mathrm{J}\\ =\frac{1}{24}{I}_B{w}_{0/A}^2+2400\mathrm{J}\\ =\frac{1}{24}\left(3I_B\right)w_{0/A}^2+2400\mathrm{J}\\ =\frac{1}{8}{I}_B{w}_{0/A}^2+2400\mathrm{J}\end{array}$

Since, role="math" localid="1668057419817" $k_{0/A/max}=\frac{1}{2}{I}_B{W^2}_{0/A/max}$then,

$\begin{array}{r}\frac{1}{2}{I}_B{w}_{0/A}^2=\frac{1}{8}{I}_B{w}_{0/A}^2+2400\mathrm{J}\\ \Rightarrow \frac{3}{8}{I}_B{w}_{0/A}^2=2400\mathrm{J}\end{array}$

Thus, role="math" localid="1668057517014" $\begin{array}{r}\frac{1}{2}{I}_B{w}_{0/A}^2=\frac{\left({\displaystyle \frac{1}{2}}\right)\left(2400\right)\left(8\right)}{3}\\ \\ \end{array}$

= 3200 J

Therefore, the maximum value of the original kinetic energy is 3200 J.