What Is the Tangential Acceleration of a Chain Sliding Down a Sphere?

randommanonea
Messages
7
Reaction score
0
A chain of length 'l' and mass 'm' lies on the surface of a smooth sphere of radius 'R' > 'l', with one end tied to the top of the sphere.

(a) Find the gravitational potential energy of the chain with reference level at the center of the sphere.

(b) Find the tangential acceleration dv/dt of the chain when the chain starts sliding down.

----------------------------------------------------------------------------------------

I was able to do the (a) part, which is a matter of simple integration and my answer came out to be {m R^2 g sin(l/R)}/l

Can someone please help me out with the (b) part.
 
Physics news on Phys.org
Hi Randommanonea,
Welcome to PF!

To find dv/dt, try to use the principle of conservation of energy i.e. find out the kinetic energy when the chain slides by an angle say ϑ.
Then differentiate the equation to get an expression for dv/dt, use v=rdϑ/dt for simplification (assuming that the sphere does not rotate).

Feel free to ask any doubt in the above steps.
 
I got my answer as :

Rg{1-cos(l/R)}/l


Is it correct ?
 
Yes its correct :approve:
 
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