- #1
alba
- 140
- 4
I am trying to learn to calculate collisions on the center of percussion of a rod.
Can you tell me how to find the moment of inertia?
Suppose we have a rod of length 1m and mass M = 12 kg., we know that ##I = 12 * 1^2 / 3 = 4## , if a collision takes place at the CoP ( at 0.166666 m from the Center of mass and 0.66666 from the pivot), is still I = 4? .
If a collision takes place at the CoM, isn't it just an ordinary collision with a body of ( I =) M = 12 ?
If an incoming ball (m = 1, v = p = 20) hits the rod at the CoM it will rebound at v = 16.92 and the rod will move at v = 3.077, without any rotation, right?
If it his the CoP, is its angular momentum = 20 * 0.6666 or 20 * 0.16666? And what is the angular momentum of the swinging rod?
Thanks for your help
Can you tell me how to find the moment of inertia?
Suppose we have a rod of length 1m and mass M = 12 kg., we know that ##I = 12 * 1^2 / 3 = 4## , if a collision takes place at the CoP ( at 0.166666 m from the Center of mass and 0.66666 from the pivot), is still I = 4? .
If a collision takes place at the CoM, isn't it just an ordinary collision with a body of ( I =) M = 12 ?
If an incoming ball (m = 1, v = p = 20) hits the rod at the CoM it will rebound at v = 16.92 and the rod will move at v = 3.077, without any rotation, right?
If it his the CoP, is its angular momentum = 20 * 0.6666 or 20 * 0.16666? And what is the angular momentum of the swinging rod?
Thanks for your help