Solving the Stuck Ring Problem Using Expansion Coefficients

  • Thread starter Thread starter skwerl
  • Start date Start date
  • Tags Tags
    Ring Stuck
AI Thread Summary
To separate a brass ring with a diameter of 10.00 cm from an aluminum rod with a diameter of 10.01 cm, the temperature must be lowered to a point where both metals contract sufficiently. The coefficients of linear expansion for brass and aluminum are 19x10^-6 and 24x10^-6, respectively. Calculating the necessary temperature involves using the formula L = L0(1 + α ΔT) to find when both diameters equalize. If the aluminum rod's diameter increases to 10.03 cm, the required temperature for separation will be different and must be recalculated. The feasibility of achieving these temperatures depends on the maximum allowable ΔT for the materials.
skwerl
Messages
4
Reaction score
0
A brass ring of diameter 10.00 cm at 19.5C is heated and slipped over an aluminum rod of diamter 10.01 cm at 19.5C. Assume the average coefficients of linear expansion are constant.

a) To what temperature must this combination be cooled to separate them? is this possible?
b) If the aluminum rod were 10.03 cm in diameter, what would be the required temperature? Is this possible?

B(brass) = 19x10^-6
B(alum) = 24x10^-6

Well...let's see... first I thought of heating the brass until it was wide enough to slip over the tube, and then I tried to set up some kind of relationship and failed miserably.

oh and by the way I'm new! hi guys!
 
Last edited:
Physics news on Phys.org
skwerl said:
Well...let's see... first I thought of heating the brass until it was wide enough to slip over the tube, and then I tried to set up some kind of relationship and failed miserably.
Welcome to Physics Forums!

When the temperature is reduced, both metals will contract. The question is will they contract enough to separate?

The length is given by: L = L0(1 + α ΔT). Set it up and solve for the temperature that will make both metals the same diameter.

Hints: You know the "unstressed" diameter of both metals at 19.5°C, so start there. Also, what is the maximum possible ΔT?
 
neat. so much more elegant than what I thought of

thanks!
 
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