How Do You Calculate Tension and Speed in Rotating Bead Physics Problems?

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SUMMARY

The discussion focuses on calculating the tension and speed of a bead sliding along a string in a rotating system. A 100 g bead on an 80 cm string, with endpoints attached to a vertical pole 40 cm apart, requires the application of Newton's second law (F=ma) and centripetal acceleration (a=v^2/r) to solve the problem. The solution involves using the properties of a right triangle to determine the horizontal distance, leading to the calculation of tension and speed at point B.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of centripetal acceleration (a=v^2/r)
  • Familiarity with right triangle properties
  • Ability to solve algebraic equations
NEXT STEPS
  • Study the application of Newton's laws in rotational motion
  • Learn about centripetal force and its calculations
  • Explore the properties of right triangles in physics problems
  • Practice solving similar bead-on-string dynamics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking examples of tension and speed calculations in practical scenarios.

Joelseph
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Homework Statement


A 100 g bead is free to slide along an 80 cm long piece of string ABC. The
ends of the string are attached to a vertical pole at A and C, which are 40
cm apart. When the pole is rotated about its axis, AB becomes horizontal.
(a) Find the tension in the string. (b) Find the speed of the bead at B.


Homework Equations


F=ma
a=v^2/r



The Attempt at a Solution


I have drawn a free body diagram, and it is a right angle triangle with the vertical side being 40cm, and the other 2 sides being 80 in TOTAL but I'm not sure how to figure out the angles or sides with that little information.

Please assist. Once I have that information, the tension of the string is the force in the horizontal direction, and the speed will be easy enough to find as well.

Thanks
 
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If x is the length of the horizontal side then X^2 + (40)^2 = (80 - x )^2
Solve for x.
 
Gotcha, thanks for the help that was perfect.
The classic 3-4-5 triangle!
 

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