Question 10.6: A homogeneous mechanical system consists of a flywheel and c...
A homogeneous mechanical system consists of a flywheel and clutch plate. The flywheel rotates in an inertial frame \Phi with angular speed \omega_f about its axis of symmetry with respect to which its moment of inertia is I_f. The clutch plate, initially at rest in Φ, has coaxial symmetry about which its moment of inertia is I_c, The flat surface of the clutch plate suddenly engages the plane surface of the flywheel and immediately adheres to it without slipping. Ignore frictional and other extraneous loads due to bearings and other attachments. Determine the angular speed of the clutch and flywheel assembly at the moment after engagement.
The clutch and flywheel rotate about a common axis of symmetry e_3, fixed in an inertial frame . We neglect extraneous loads and frictional effects of attached devices, and recall (10 .75) and the instantaneous torque-impulse principle (10.55). The clutch plate is at rest initially; hence, the instantaneous torque-impulse exerted on the clutch c by the flywheel f about a point Q on the axis and in their plane of contact is \mathscr{M}_{Q c f}^*=\Delta \mathbf{h}_{Q c}=I_c \omega_c \mathbf{e}_3 where \boldsymbol{\omega}_c=\omega_c \mathbf{e}_3 is the angular velocity of the clutch after its engagement.The instantaneous torque-impulse exerted on the flywheel by the clutch is \boldsymbol{W}_{Q f c}^*=\Delta \mathbf{h}_{Q f}=I_f\left(\Omega_f-\omega_f\right) \mathbf{e}_3 . The adherence condition requires that the final angular speed of the flywheel \Omega_f=\omega_c, as the two plates instantaneously rotate together. Because the torques are mutual internal torques, \mathscr{M}_{Q c f}^*=-\mathscr{M}_{Q f c}^* Therefore, at the impulsive instant, I_c \omega_c=-I_f\left(\omega_c – \omega_f\right), and hence the angular speed of the clutch and flywheel assembly is
\mathbf{h}_{r Q}=I_{33} \omega(t) \mathbf{e}_3=I_Q \boldsymbol{\omega}, \quad \mathbf{M}_Q=I_{33} \dot{\omega} \mathbf{e}_3=I_Q \dot{\boldsymbol{\omega}} (10 .75)
\mathscr{M}_Q^* \equiv \operatorname{limit}_{t \rightarrow t_0} \mathscr{M}_Q\left(t ; t_0\right) (10.55)
\omega_c=\frac{I_f}{I_c + I_f} \omega_f.
Clearly, \omega_c<\omega_f, as one may expect intuitively. The flywheel, however, because of its greater mass and size, usually has a much greater moment of inertia so that I_c / I_f \ll 1, and hence \omega_c=\omega_f, very nearly.
Alternatively, because there are no external impulsive forces or torques, the total torque-impulse \mathscr{M}_Q^* about the axis of rotation must vanish, and hence \Delta \mathbf{h}_Q=0 at the impulsive instant, that is, the total moment of momentum about the axis \mathbf{e}_3 is constant in the inertial frame. Thus, the final moment of momentum of the system equals its initial moment of momentum: \left(I_f + I_c\right) \omega_c \mathbf{e}_3=I_f \omega_f \mathbf{e}_3. This yields the same result given above.