Rotating System in an Inelastic Collision

AI Thread Summary
The discussion focuses on calculating the critical angular speed at which a string breaks in a rotating system involving two masses and a hollow cylinder. The tension in the string is identified as the centripetal force that becomes insufficient when the angular speed exceeds a certain threshold. Participants clarify that the string's breaking point is due to the inability to provide the necessary centripetal acceleration as the system rotates faster. There is confusion regarding the application of the centripetal force equation, leading to questions about potential miscalculations. The conversation emphasizes the importance of correctly applying the principles of angular momentum and centripetal force in solving the problem.
IWuvTeTwis
Messages
7
Reaction score
0

Homework Statement


Suppose that in the figure below, script i = 0.88 m, L = 2.2 m, M = 1.2 kg, and m = 0.6 kg. The string breaks when the system's angular speed approaches the critical angular speed ωi, at which time the tension in the string is 108 N. The masses then move radially outward until they undergo perfectly inelastic collisions with the ends of the cylinder. Assume that the inside walls of the cylinder are frictionless. (For clarification, M = 2 multiplied by m and the moment of inertia of the hollow cylinder is ML^2/10. Consider the sliding masses to be point masses.)

Find the critical speed that requires the string to break. Also find the final speed after the inelastic collision.

Homework Equations


Conservation of Angular Momentum: Linitial = Lfinal
L = I*omega

The Attempt at a Solution


Because of the law of conservation of momentum I realize that I can forge a relationship between the angular speed before and after the string breaks. Iinitial*omegainitial = Ifinal*omegafinal

However, what I am unsure of is what force causes the string to break. I don't think its the centripetal force since it pushes inwards. Can it be caused by some torque? I would be welcome any suggestions.
 

Attachments

  • Rotating System.gif
    Rotating System.gif
    15.7 KB · Views: 591
Physics news on Phys.org
I think you're not understanding centripetal force. Centripetal force is the force that the string applies to keep the mass moving in a circle. In other words, it's the tension in the string. When the system is rotating too fast, the mass's centripetal acceleration becomes too high. The string tries to supply the necessary force to provide this acceleration, but it isn't strong enough, so it breaks.
 
Ok, I've tried using the equation F = 2*m*r*omega^2 where r = 1/2i. The 2 is there because there are two masses. However, I get an answer of 50.3 1/s^2 which doesn't seem to be right. Am I misapplying the centripetal force? Or is there another force I'm not accounting for?
 
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...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top