Energy and momentum - pendulums

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SUMMARY

The discussion focuses on the physics of a pendulum, specifically analyzing the motion of a heavy ball swinging on a string with a radius of 1 meter. The key equations used include the conservation of mechanical energy, represented by mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2. The ball's speed at the lowest point P is derived from gravitational potential energy converting to kinetic energy, while the speed at the highest points Q and Q' involves calculating the height difference due to the angle of 20 degrees from the vertical. The acceleration at both points is determined using the gravitational force and the geometry of the pendulum's motion.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with gravitational potential energy (PE) and kinetic energy (KE) equations
  • Basic trigonometry for calculating height changes using angles
  • Knowledge of Newton's laws of motion, particularly regarding acceleration
NEXT STEPS
  • Calculate the ball's speed at point P using mgh = 1/2mv^2
  • Determine the ball's acceleration at point P using a = g
  • Analyze the energy transformation at points Q and Q' using the conservation of energy equation
  • Explore the effects of air resistance on pendulum motion for a more advanced understanding
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of pendulum dynamics in classroom discussions.

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Homework Statement



A heavy ball swings on a string in a circular arc of radius 1m.
The two highest points of the ball's trajectory are Q and Q'; at these points the string is +/- 20 degrees from the vertical. Point P is the lowest point of the ball's trajectory where the string hangs vertically down. The acceleration of gravity is 9.8 m/s^2.

1.) What is the ball's speed at the point P? Neglect air resistance and other frictional forces.
2.) What is the magnitude of the ball's acceleration at the point P?
3.) What is the ball's speed at the point Q?
4.) What is the magnitude of the ball's acceleration at the point Q?

Homework Equations



mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2

The Attempt at a Solution



9.8(1-cos(20)) + 1/2v1^2 = 0 + 1/2v2^2
0.591 + 1/2v1^2 = 1/2v2^2

How can I solve this with two unknowns?
 
Physics news on Phys.org
Why do you think there are 2 velocities to consider?

PE at highest = KE at lowest

m*g*h = m*v2/2
 

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