Mechanical Energy: Solving for Speed of Ball at Lowest Point

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SUMMARY

The discussion focuses on calculating the speed of a ball at the lowest point of a vertical circular motion using principles of mechanical energy. Given a thin rod of length L = 2.20 m and an angle of release θ = 19.0°, the mass of the ball is m = 500 kg. The relevant equations include work done W = mgd(cos θ) and kinetic energy KE = (1/2)mv². The solution involves determining the work done by gravitational force and equating it to the kinetic energy to find the velocity at the lowest point.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts
  • Familiarity with the equations of motion in physics
  • Knowledge of trigonometric functions, specifically cosine
  • Ability to manipulate algebraic equations for solving variables
NEXT STEPS
  • Study the conservation of mechanical energy principles in physics
  • Learn how to derive velocity from potential energy in circular motion
  • Explore the role of angles in calculating work done in physics
  • Investigate the effects of mass and length on the speed of objects in rotational motion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of rotational motion and energy conservation principles.

norcal
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Homework Statement



A thin rod, of length L and negligible mass, that can pivot about one end to rotate in a vertical circle. A heavy ball of mass m is attached to the other end. The rod is pulled aside through an angle and released.
What is the speed of the ball at the lowest point if L = 2.20 m, = 19.0°, and m = 500 kg?

Homework Equations



W=mgd(cos theta)
KE=(1/2)mv^2

The Attempt at a Solution



How do I get the velocity? Can I get it by using the above equations if I solve for W? Where do I plug the answer for W into find the velocity?
 
Physics news on Phys.org
Do you know what W is as defined in your post ? Can you give it a name ?
 
W=work done
 

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