An application of Conservation of Angular Momentum One of the applications of the law of conservation of angular momentum involves a helicopter with a single propeller. A/c the book, a helicopter with one propeller would rotate itself in the opposite direction. However, I am not able to visualise this phenomenon. Can you please explain this, preferably with the help of a figure?
 A: Consider the two phases of motion of the blade of the helicopter. First their angular velocity will increase and then become constant. To produce angular acceleration torque is required. Torque is also required to maintain constant angular velocity otherwise friction will slow it down. Applying anti-clockwise torque would result in clockwise moment generation on the helicopter(except its blades) thus it will start rotating in the opposite direction to the blade.
To escape applying torque analysis on each body we apply angular momentum conservation on the whole system about an axis/point wherever external torque is zero.
A: Here is a simple analogy.  Assume you are standing upright, stationary on ice (frictionless surface).  The external forces (gravity and constraint from surface) provide no torque about your center of mass (CM), so the angular momentum about your CM is constant and zero.  If you swing your arms to the left your body rotates to the right to keep the angular momentum zero.
For the helicopter the propeller is the "arms" and the body of the helicopter is the "body".
Hope this helps.
A: As the helicopter applies torque to its top rotor, the rotor is applying an equal and opposite torque to the helicopter. The helicopter and the top rotor would be spinning in opposite directions without the tail rotor to push against this equal and opposite  torque applied by the top rotor to the helicopter.
A: Conservation of angular momentum. Since no external torques are applied to the helicopter (the forces causing the torques that spin the rotor are internal), the total angular momentum must stay constant, and equal to zero (since initially there is no angular speed). In other words, the following must be true $$I_\mathrm{blade}\omega_\mathrm{blade} + I_\mathrm{heli}\omega_\mathrm{heli}=0$$
Therefore, if $\omega_\mathrm{blade}$ is positive, then $\omega_\mathrm{heli}$ will be negative, i.e. the helicopter must spin in the opposite direction.
