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## Q. 10.17

A heavy flywheel in the form of a solid circular disk of radius R is attached by a bearing to the end of a simply supported beam of length L and constant flexural rigidity EI as shown in Figure 10.31. The flywheel has weight W and rotates at constant angular velocity ω. If the bearing suddenly freezes, what will be the reaction P at support A? Neglect the mass of the beam itself.

## Verified Solution

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

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