Calculate Tension R of Mass m on String of Length l at Temp T

  • Thread starter Thread starter pt176900
  • Start date Start date
  • Tags Tags
    Stat mech
AI Thread Summary
The discussion centers on calculating the tension R in a string of length l with a mass m rotating at temperature T. The tension's x-components cancel out, leaving the y-components to provide the necessary centripetal force. The energy of the rotating mass is expressed as 1/2 I w^2, equating to the average thermal energy of 1/2 kT due to thermal equilibrium. However, there is confusion about transitioning from energy to tension, especially considering the massless assumption of the string. The impact of temperature on the string's length through thermal expansion is also questioned, highlighting the complexity of relating macroscopic and microscopic energy systems.
pt176900
Messages
27
Reaction score
0
A weight of mass m is fixed to the middle point of a string of length l and rotates about an axis joining the ends of the string. The system is in contact with its environment at temperature T. Calculate the tension R between the ends of the string in terms of its dependence upon distance x between the ends and the temperature.

My thoughts:

clearly, the x-components of the tension will cancel and we are left with the y-components of the tension which provides the centripital force.

The energy of the rotating mass is given by 1/2 I w^2 (where w is the angular frequency). Since the string is in thermal equilibrium with the envrionment we can equate it with the average thermal energy for 1 degree of freedom: 1/2 kT

What I don't understand, is how to go from the energy to the tension on the string.
 
Physics news on Phys.org
Well you know the energy of the string plus the mass. Since the string is assumed to be massless(right?) then you know the velocity as mass rotates.

The distance between the two ends of the string X will decide the radius at which the mass rotates.

Does this help?
 
I'm very sure u can't equate anything macroscopic with \frac{1}{2} kT

Daniel.
 
Is there a way to solve the problem if it is macroscopic?
 
dextercioby said:
I'm very sure u can't equate anything macroscopic with \frac{1}{2} kT

Daniel.

even if you did, if a macroscopic energy system had kT worth of energy, ie. 0.026 eV at room temperature, it'd be basically zero relative to the moment of inertia of the mass.

is there a coefficient of thermal expansion for the string or something like that, which would affect the length of the string as a function of temperature?
 
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