Solving Properties of Matter Problem: Steel Ring Expansion Force

  • Thread starter Thread starter Amith2006
  • Start date Start date
  • Tags Tags
    Matter Properties
Amith2006
Messages
416
Reaction score
2
Sir,
Please help me in solving this problem.
# A steel ring of radius r and cross sectional area A is fitted on to a wooden disc of radius R(R>r). If the Young’s modulus be Y, then what is the force with which the steel ring expands?
I just don’t have any idea on how to go about this problem.
 
Physics news on Phys.org
Is that the exact wording of the question ?

It is extremely ambiguous, and as such, is bordering on nonsensical.
 
For a very small portion of the ring (dl=Rd \theta) you have the static equilibrium condition:
2 T sin(\frac{d \theta}{2})=N
where N is the force between the ring and the cylinder. Now you use
sin (\frac{d \theta}{2}) ~= \frac{d \theta}{2}
and T from the Young law for the small element dl (when it expands between dl_0=r d \theta and dl=R d\theta) and so on...
 
Last edited:
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top