
#1
May908, 11:28 AM

P: 414

1. The problem statement, all variables and given/known data
Q] A meter stick weighing 240 grams is pivoted at it's upper end in such a way that it can freely rotate through this end. A particle of mass 100 gms is attached to the upper end of the stick to a string of length 1m. Initially the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise. 3. The attempt at a solution This is a case of pure inelastic collision. In this case: [tex] \Delta {mv} = 0 [/tex] The potential energy of the particle at the position in rest is 1J. Hence, when it reaches the bottom, it's P.E will get converted to K.E. Hence: [tex] \frac{1}{2}mv^2 = 1 [/tex] and hence, [tex] v = \sqrt{2}{m} = \sqrt{20} = 2\sqrt{5} [/tex] and the momentum is given by: [tex] p = mv = 0.2\sqrt{5} [/tex] Now, when it collides with the rod below, the momentum of the whole system at this point will be the same, but the mass of the system is now 0.24 + 0.1 = 0.34 gms. Hence, the velocity of the system at this point is given as: [tex] v = \frac{0.2\sqrt{5}}{0.34} [/tex] Now, the situation can be simplified to: There is a system of a bob attached to a rod, with: [tex] I = 0.24~ kgm^2~~;~~m = 0.34 ~kg [/tex] A gravitational force acts on it and i am left to determine by how much the system will be displaced [angular] under this force. I have no idea how to do that as of now. Any help is appreciated. 



#2
May908, 11:39 AM

Mentor
P: 40,878

Comments:
(1) During the collision: Since the stick is constrained by its pivot, translational momentum is not conserved. But what is? (2) After the collision: Energy is conserved as the stick+mass swings up. Hint: Where's the center of mass? 



#3
May908, 11:54 AM

P: 414





#5
May908, 12:23 PM

P: 414

I tried doing it using conservation of angular momentum now. Just before the bob strikes the rod, it's angular momentum is given by:
[tex] l = mvr= (0.1)(2\sqrt{5})(1) = 0.2\sqrt{5} [/tex] After collision, the angular momentum of the system will remain constant. Hence, [tex] I\omega = 0.2\sqrt{5} [/tex] [tex] \omega = \sqrt{5}{1.2} [/tex] The K.E. of the system at this point is given as: [tex] K.E = \frac{1}{2}I\omega^2 = \frac{10}{24} [/tex] The center of mass is at a distance 'r' from the pivoted end: [tex] r = \frac{1}{0.34}[(0.24)(0.5) + (0.1)(1)} = \frac{11}{17} [/tex] At this point, the gravitational force will be acting. The perpendicular distance of this point from where we computed the K.E is given [or took the P.E to be zero] is given by: [tex] h = 1  \frac{11}{17}cos(\theta) [/tex] On equating the GPE at this point with the K.E found earlier, i am getting the theta as ArcCos[0.9], whereas i should be getting ArcCos[0.75]. What am i doing wrong here? 



#7
Aug2908, 05:51 AM

P: 1

how we can find the moment of inertia of aflywheel peactically , you don,t know the mass of flywheel and also consider friction?



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