Question

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system like the one in (Figure 1). 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 ω0; 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 3100 J of thermal energy to be developed when the connection is made.

BEFORE TO Forces and are along the axis of rotation and thus exertne torque about this axis on cither disk AFTER I Als

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?

Answer 1

Conservation of Angular momentum:

Angular momentum is defined as the cross product of the moment of inertia and the angular velocity of the object. If the external torque on the system is zero then the angular momentum of the system remains conserved.

There are no external torques to the system, so angular momentum is conserved, so the final angular speed $\omega$ satisfies:

(IA +IBw= IAWO

It is given that:

$I_A =\frac{1}{3}I_B$

Therefore,

$\omega =\frac{1}{4}\omega _{0}$

Now consider the kinetic energies before and after coupling.

$K_{after} =\frac{1}{2}( I_A + I_B ) \omega^{2} = \frac{I_A \omega _{0}^{2}}{8}$

$K_{before} = \frac{I_A \omega _{0}^{2}}{2}$

Therefore the loss in kinetic energy is:

$\frac{3}{8}I_A \omega _{0}^{2}$

This went into heat, and since we want that maximized to Qmax (3100J in this problem) we have

$\frac{3}{8}I_A \omega _{0}^{2}<Q_{max}\implies \frac{1}{2}I_A \omega _{0}^{2}<\frac{4}{3}Q_{max}$

So the maximum initial kinetic energy of A is :

$\frac{4}{3}Q_{max}\implies \frac{4}{3}(3100J)=\mathbf{4133J}$