Falling Fun
A weight sitting on top of a dropped tennis racket will fall with it.
What if the stringed end of the racket has a rod running through it so that the racket will swing rather than fall freely? Will the weight still remain on top of the racket?
Follow up questions (watch the video first). Is the end of the racket falling faster than free fall or is the weight falling at less than free fall? How can this happen?
Comments
Follow Up Question 1: The end of the racket must be falling faster, because the acceleration due to gravity is a constant, so the weight couldn't possibly be falling slower.
Follow Up Question 2: I'm not sure. some of the energy from the string end of the racket must be transferred to the handle through angular momentum, making it fall faster? I'm just making things up at this point.
It is indeed going faster than free fall (initially at least). If you analyze the torque due to gravity on the center of mass, it turns out that you get an initial accel=3/2*g/L*r where r is the distance out from the pivot point and L is the length of the racket (approximating the racket as a uniform rod). That means the end of the rod accelerates at 3/2*g, the center at 3/4*g, and a=g at 3/4*L.
Conceptually I would say that the faster than g acceleration is due to the fact that the pivot point exerts an upward force which causes it to rotate. Imagine a rod (or racket) that is in free fall and then one end collides with some object (say a table). This would send the rod spinning with a similar effect.