Circular motion with a string with weight

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SUMMARY

The discussion focuses on calculating the tension in a cord of mass m1 attached to a block of mass m2, moving in a horizontal circle on a frictionless table. The problem requires setting up an integral equation to find the tension as a function of radial position along the cord, defined from 0 to L. The key equations involved include centripetal acceleration, expressed as a = v^2/r, and the force equation F = ma. The participants emphasize the importance of considering the mass of the cord and its linear mass density, λ = m1/L, in the calculations.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with integral calculus
  • Knowledge of centripetal acceleration concepts
  • Basic principles of tension in strings with mass
NEXT STEPS
  • Study the derivation of tension in non-massless strings
  • Learn about linear mass density and its applications in physics
  • Explore integral calculus techniques for solving physics problems
  • Investigate the effects of mass distribution on tension in circular motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for examples of tension in systems with mass. This discussion is also beneficial for anyone preparing for advanced physics problems involving integrals and dynamics.

giggidygigg
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Homework Statement


A block of mass m2 is attached to a cord of mass m1 and length L, which is fixed at one end. (Note, not a massless cord!) The block moves in a horizontal circle on a frictionless table. If the period of the circular motion is P, find the tension in the cord as a functions of radial position along the cord, 0 <_ r <_ L. (<_ meaning greater than or equal to)


Homework Equations


a = v^2/r, F = ma


The Attempt at a Solution


So I could immediately tell that this problem involves setting up an integral equation of some sort. I've learned in class how to solve for integral equations when a block is hanging on a ceiling or things as simple as that, but none of these.
So if L is the distance from the center of the circle to some arbitrary point on the cord, I think the tension at the very end of the cord connected to the mass will just be the centripetal acceleration of the mass multiplied by m1, so there's one of the limits of the integral equation. The other limits should be 0 and L for the radius. The last one will be the unknown for which I have to solve for.
But when I set up my equation and solved for it, I somehow got T(L) = m_1*2*pi*L + T(0), which doesn't make sense.
 

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giggidygigg said:
So if L is the distance from the center of the circle to some arbitrary point on the cord, I think the tension at the very end of the cord connected to the mass will just be the centripetal acceleration of the mass multiplied by m1

Why? Does the mass of the block disappear? For any point on the string at a distance x, if you assume the linear mass density to be \lambda =\frac{m_1}{L} then the string at that point is \lambda xdx+ m_2. Figure out the variation of x and then integrate.
 
Last edited:
oops i meant m2. but anyway i'll try to carry it out with your advice
 

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