- #1
alba
- 140
- 4
I read the following example in a thread in this forum:
A ball (m = 1 Kg , v = p =+22 m/s, Lm = +11, Ke = 242 J) hits the tip of a rod (M = 10Kg , length = 1m, i = 5/6 ).
the ball bounces back with v, p = -11.846 m/s , L = -5.923, (Ke = 70.16) the rod translates with v = 3.3846m/s , p = 33.846 , (Ke = 57.2J) and rotates about its CoM with ##\omega## = 16.58 (Ke = 114.556).
Ke is conserved : 70.16 + 57.2 + 114.6 = 242
linear momentum is conserved: 33.846 - 11.846 = 22
L was +11
after the impact we have
Lm = -5.923
Lr = 16.58 (##\omega## * i ) 5/6 = 13.82
13.82 - 5.92 = 7.9
It seems that angular momentum is not conserved. Did I go wrong somewhere?
A ball (m = 1 Kg , v = p =+22 m/s, Lm = +11, Ke = 242 J) hits the tip of a rod (M = 10Kg , length = 1m, i = 5/6 ).
the ball bounces back with v, p = -11.846 m/s , L = -5.923, (Ke = 70.16) the rod translates with v = 3.3846m/s , p = 33.846 , (Ke = 57.2J) and rotates about its CoM with ##\omega## = 16.58 (Ke = 114.556).
Ke is conserved : 70.16 + 57.2 + 114.6 = 242
linear momentum is conserved: 33.846 - 11.846 = 22
L was +11
after the impact we have
Lm = -5.923
Lr = 16.58 (##\omega## * i ) 5/6 = 13.82
13.82 - 5.92 = 7.9
It seems that angular momentum is not conserved. Did I go wrong somewhere?