Just got owned on this exam question

  • Thread starter Thread starter anotherghost
  • Start date Start date
  • Tags Tags
    Exam
AI Thread Summary
The discussion revolves around a physics exam question involving two masses, m1 and m2, connected by a rope over a frictionless pulley. The problem requires calculating the acceleration of m1, the tensions in the rope, the angular velocity of the pulley, and the inertia of the pulley, given that m2 descends 2 meters in 5 seconds. Participants suggest using free-body diagrams and Newton's second law rather than conservation of energy to find acceleration. One contributor realizes they can derive the necessary values once they determine the acceleration. The conversation highlights the importance of understanding the relationship between linear and angular motion in such problems.
anotherghost
Messages
10
Reaction score
0

Homework Statement



You've got two masses, m1 and m2, hanging from opposite ends of a rope that goes over a pulley with radius R. Both the masses are suspended by the rope alone and the pulley is on a frictionless peg. M2 moves down by x2 meters in 5 seconds. I believe that's all the variables we were given - m1, m2, radius R, and 2 meters in 5 seconds. EDIT: The pulley doesn't slip.

Now find the accelleration of m1, the tensions of the rope on each side of the pulley, the angular velocity of the pulley, and the inertia of the pulley.

Homework Equations


The Attempt at a Solution



I have no idea how to do this without being given the mass of the pulley or its inertia.

Started out with m1gx + m2gx = (1/2)m1v^2 + (1/2)m2v^2 + (1/2)I(V/R)^2 + m1g*2x.
 
Physics news on Phys.org
Since the question asks for acceleration, you probably don't want to use conservation of energy because conservation of energy equations have no acceleration terms. Instead, use plain old free-body diagrams and Newton's second law.

Since the acceleration is constant and d=v0t+(1/2)at^2, you can calculate acceleration with the fact that m2 dropped by 5 m in 2 s. See where this takes you.
 
Oh man, duh. I can't believe I didn't see that. I can get the rest of it if I have that a no problem. Guess I'll just have to chalk that one up to a brain fart, hope I'll still get partial credit. :)
 
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