Use physics to explain the reason a boomerang returns to its thrower.

Answer 1

First, we need to understand that a boomerang is an airfoil; it produces lift, and that is why it stays airborne. The lift equation is Lift = 1/2 * coefficient of lift * fluid density * Area of the lifting surface * relative air velocity ². The coefficient of lift is a property of the airfoil and something called the angle of attack, which we will assume is constant to simplify the problem. Likewise, fluid density and area are assumed to be constant.

That leaves relative air velocity. Relative air velocity is defined as the velocity of air as seen from a reference point. Let's say an aircraft is moving 300 km/h into a 100 km/h head wind. The relative air velocity as seen from the aircraft would be 300 + 100 = 400 km/h.

The relative air velocity over the boomerang is neither uniform nor symmetric. Let's look at two points; one at each end of the boomerang. One end will spinning away from the thrower while the other end spins toward from the thrower. This means that one end is experiencing a relative air velocity higher than that of the center of rotation while the other end experiences reduced relative air velocity. This applies down the entire length of the boomerang; the side that is spinning away from the thrower will experience more lift than the side that is moving towards the thrower by the equation given earlier.

This causes the boomerang to start to tilt with the side spinning forward tilting up, and it turns for the same reason an airplane does when it rolls - now the direction of lift isn't straight up; it's partially to the side. Eventually, the boomerang turns around, and an experienced thrower can make the boomerang return to its release point, but that requires practice.