Asked by Nika
A thin stick of mass M = 3.1 kg and length L = 1.6 m is hinged at the top. A piece of clay, mass m = 0.7 kg and velocity V = 4.0 m/s hits the stick a distance x = 1.20 m from the hinge and sticks to it.
1. What is the angular velocity of the stick immediately after the collision?
2. What is the ratio of the final mechanical energy to the initial mechanical energy?
1. What is the angular velocity of the stick immediately after the collision?
2. What is the ratio of the final mechanical energy to the initial mechanical energy?
Answers
Answered by
Damon
angular momentum around pivot is the same before and after
before
angular momentum = m v r = .7*4*1.2
= 3.36 kg m^2/s
after let w = omega = ang velocity about pivot
angular momentum = I w
I = (1/3)(3.1)(1.6^2) + .7(1.2^2)
= 2.645 + 1.008 = 3.65
so
I w = 3.36
3.65 w = 3.36
w = .921 radians/second
part 2
before
(1/2) m v^2 = (1/2).7(16) = 5.6 Joules
after
(1/2) I w^2 = (1/2)(3.65)(.921^2) = 1.55 Joules
1.55/5.6 = .276
before
angular momentum = m v r = .7*4*1.2
= 3.36 kg m^2/s
after let w = omega = ang velocity about pivot
angular momentum = I w
I = (1/3)(3.1)(1.6^2) + .7(1.2^2)
= 2.645 + 1.008 = 3.65
so
I w = 3.36
3.65 w = 3.36
w = .921 radians/second
part 2
before
(1/2) m v^2 = (1/2).7(16) = 5.6 Joules
after
(1/2) I w^2 = (1/2)(3.65)(.921^2) = 1.55 Joules
1.55/5.6 = .276
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.