Question 10.17: A heavy flywheel in the form of a solid circular disk of rad...

We know for solid circular cylinder, the mass moment of inertia about its centroidal axis is

\bar{I}=\frac{1}{2} M R^2=\frac{W}{2 g} R^2

and its rotational kinetic energy is T=\bar{I} \omega^2 / 2, \text { or }

T=\frac{W}{4 g} \omega^2 R^2

Ignoring any energy loss, we can say change in kinetic energy, ΔT is

\Delta T=\frac{1}{2} \bar{I} \omega^2=\frac{W}{4 g} \omega^2 r^2

which is the strain energy stored within the beam. Now,

\Delta T=U_{\text {bending }}=\left\lgroup\frac{1}{2 E I} \right\rgroup \int_0^L M_x^2 d x

or            \left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^L M_x^2 d x=\left\lgroup \frac{W}{4 g} \right\rgroup \omega^2 R^2              (1)

M_x can be found from the following free-body diagram of the beam as shown in Figure 10.32.

From Figure 10.32(b), we note M_x=Px and from Eq. (1)

\left\lgroup \frac{1}{2 E I}\right\rgroup \int_0^L P^2 x^2 d x=\left\lgroup \frac{W}{4 g}\right\rgroup \omega^2 R^2

or            \frac{P^2 L^3}{6 E I}=\frac{W}{4 g} \omega^2 R^2

or            P^2=\frac{6}{4} \frac{(W E I)\left(\omega^2 R^2\right)}{g L^3}

\Rightarrow P=\sqrt{\frac{3}{2} \frac{(W E I)\left(\omega^2 R^2\right)}{g L^3}}

Therefore, the required reaction at left-end support is

P=\sqrt{\frac{3 E I W \omega^2 R^2}{2 g L^3}}