Angular Velocity and Tension Around a Fixed Cylinder

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SUMMARY

The discussion focuses on the dynamics of a particle of mass m attached to a light string that wraps around a fixed vertical cylinder of radius a. The initial angular velocity of the cord is denoted as w0 when the distance from the particle to the contact point is b. After the cord rotates through an additional angle theta, the angular velocity is expressed as w = w0 / (1 - (a/b)theta), and the tension in the string is T = m * b * w0 * w. Key insights include the role of torque and the perpendicular nature of the force due to the string relative to the velocity vector.

PREREQUISITES
  • Understanding of angular momentum and torque
  • Familiarity with the equations of motion in circular dynamics
  • Knowledge of the relationship between linear and angular velocity
  • Basic principles of tension in strings and forces
NEXT STEPS
  • Study the principles of angular momentum conservation in non-conservative systems
  • Learn about the dynamics of particles in circular motion
  • Explore the derivation of tension in strings under varying conditions
  • Investigate the effects of angular displacement on angular velocity
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of angular motion and tension in strings.

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Homework Statement


A particle of mass m at the end of a light string wraps itself about a fixed vertical cylinder of radius a. All the motion is in the horizontal plane (disregard gravity). The angular velocity of the cord is w0 when the distance from the particle to the point of contact of the string and cylinder is b. Find the angular velocity and tension in the string after the cord has turned through an additional angle theta.

Hints from my professor:
-There is a torque acting on m so angular momentum is not conserved. But something even more simple is.
-The force on m due to the string is always perpendicular to the velocity vector.


Homework Equations


T=mv^2/r
w= d(theta)/dt
w=vsin(theta)/r
?

The answer given is:
w=w0/(1-(a/b)theta), T=m*b*w0*w



The Attempt at a Solution


I saw that v*sin(theta) = w0 so I got as far as w=w0/r but that's it. And I'm confused in the tension equation as to how one would obtain a as b*w0*w.
 
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I think the velocity remains constant. Kinetic energy constant.
At least, that assumption leads to the given answer!
 

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