How Do Kinetic and Potential Energy Differ in Physics?

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SUMMARY

The discussion clarifies the relationship between kinetic energy (KE) and potential energy (PE) in physics, specifically highlighting the equation PE + KE = constant. The kinetic energy is represented as (1/2)m(v^2), while potential energy is expressed as mgr(1 + cos(θ). At the lowest point of motion, where θ = π, potential energy is zero, and kinetic energy equals (1/2)mv₀². This establishes that the total mechanical energy remains constant throughout the motion.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) concepts
  • Familiarity with the equations of motion in physics
  • Knowledge of centripetal force and its relation to circular motion
  • Basic trigonometry, particularly the cosine function
NEXT STEPS
  • Study the conservation of mechanical energy in closed systems
  • Learn about centripetal acceleration and its mathematical formulation
  • Explore the implications of gravitational potential energy in various contexts
  • Investigate the role of angular velocity in circular motion dynamics
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of energy conservation in motion.

girlwhoneedsmathhelp
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Homework Statement
I saw this in my school textbook and have no idea how they manage to get this equation :

The context is Circular Motion :
"A particle on a string, w(angular velocity) varies throughout the motion. As you saw earlier, the value of w at any instant is given by the energy equation, which in this case is :
(1/2)m(r^2)(w^2) + mgr(1+cos(theta)) = (1/2)m(u^2) where u is the speed of the particle at the lower point."

Thank You!
Relevant Equations
1/2mv^2
mrw^2
I know that (1/2)m(u^2) is KE and initially I thought this showed PE=KE but I don't think so anymore...
I believe this has something to do with acceleration and Centripetal force but I'm so so confused
 
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I can make sense of this if the configuration is like this (##+## and ##-## represent the sign of ##\cos\theta##):
20200726_141000.jpg
 
girlwhoneedsmathhelp said:
I thought this showed PE=KE
No, it shows PE+KE=constant.
It might be clearer if we use the subscript 0 for values at the lowest point and v for the speed at any instant. So their u is my v0.
Then we have rω=v, at all times, and PE+KE=##\frac 12mr^2\omega^2 + mgr(1+\cos(\theta))=\frac 12mv^2 + mgr(1+\cos(\theta))##.
At the lowest point, ##\theta=\pi##, the PE is zero (by choice) and the KE is ##\frac 12mv_0^2##, so ##\frac 12mr^2\omega^2 + mgr(1+\cos(\theta))=\frac 12mv_0^2##.
 
Ah yes I see thank you both!
 
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