A thin uniform bar, hanging, is hit by a ball. Find the angular velocity

AI Thread Summary
A thin, uniform metal bar weighing 75N and 3m long is struck by a 2 kg clay ball traveling at 14 m/s, 1.3m below the ceiling. The problem involves calculating the angular velocity of the combined system after the collision, using the principle of conservation of angular momentum. Initially, an incorrect formula was applied, leading to an erroneous result of 1.18 rad/s. After clarification, the correct calculation yielded an angular velocity of 1.38 rad/s. The discussion highlights the importance of accurately applying physical equations in dynamics problems.
Phys121VIU
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Homework Statement


A thin, uniform metal bar, 3m long and weighing 75N, is hanging veritcally from the ceiling by a frictionless pivot. Suddenly it is struck 1.3m below the ceiling by a small 2 kg clay ball, initially traveling horizontally at 14m/s. The ball sticks to the bar

Find the angular velocity of the ball+bar just after the collision.


Homework Equations


The Attempt at a Solution



Angular momentum is conserved so my understanding is that

Angular momentum of the ball = angular momentum of the bar and bar

Lball = (Ibar + Iball2

(2kg)(14m/s)(1.3m) = [(1/3)(75N/9.8m/s2)(3m)2 + (2kg)(1.3m)22

so i worked it out and got ω = 1.18 rad/s

but that's not correct...I can't think of any other way to approach it
 
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Phys121VIU said:

Homework Statement


A thin, uniform metal bar, 3m long and weighing 75N, is hanging veritcally from the ceiling by a frictionless pivot. Suddenly it is struck 1.3m below the ceiling by a small 2 kg clay ball, initially traveling horizontally at 14m/s. The ball sticks to the bar

Find the angular velocity of the ball+bar just after the collision.


Homework Equations


The Attempt at a Solution



Angular momentum is conserved so my understanding is that

Angular momentum of the ball = angular momentum of the bar and bar

Lball = (Ibar + Iball2

Are you sure that the last formula is correct?

ehild
 
Wow..its definitely not squared..thanks:P

Found the answer..1.38rad/s!
 
Well done !:smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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