Conservation of Energy of Mass on Crane

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SUMMARY

The discussion focuses on the conservation of energy principles applied to a mass suspended from a crane. The angle of swing for the mass is determined using the potential energy equation PE = mgL(1 - cosθ), where L is the length of the cable and θ is the angle of swing. For the initial speed of the crane when the angle is 50 degrees and L = 6m, the relationship between potential energy (PE) and kinetic energy (KE) is utilized, leading to the equation (PE + KE)initial = (PE + KE)final. The mass cancels out in the equations, simplifying the calculations.

PREREQUISITES
  • Understanding of potential energy (PE) and kinetic energy (KE) concepts
  • Familiarity with trigonometric functions, specifically cosine
  • Knowledge of basic physics equations related to motion and energy conservation
  • Ability to manipulate algebraic equations to solve for unknown variables
NEXT STEPS
  • Explore the derivation of the potential energy equation PE = mgh in different contexts
  • Learn about the conservation of mechanical energy in physics
  • Investigate the effects of varying lengths (L) and angles (θ) on the dynamics of pendulum motion
  • Study the application of trigonometric identities in solving physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of these concepts in action.

elf197320501
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Homework Statement


A mass is suspended from a crane by a cable of length L. The crane and the mass is moving at constant speed V. The crane stops and the mass on the cable swings out.

What is the angle that the mass swings?
If the angle is 50 degrees and L=6m, what is the initial speed of the crane?

Homework Equations

The Attempt at a Solution


For the first part where it asks for the angle, since no numbers are given I assume they're just asking for the equation. My attempt at this is using the equation PE=mgh where h=L(1-cosθ) so it'll be PE=mgL(1-cosθ). The second part for the initial speed can be solved by (PE + KE)initial =(PE+KE)final.
 
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elf197320501 said:

The Attempt at a Solution


For the first part where it asks for the angle, since no numbers are given I assume they're just asking for the equation. My attempt at this is using the equation PE=mgh where h=L(1-cosθ) so it'll be PE=mgL(1-cosθ).
It's likely you're expected to express θ in terms of L and V (and g), m will cancel out. You'll use the same equation you will for the second part.

The second part for the initial speed can be solved by (PE + KE)initial =(PE+KE)final.
Yes, just plug and chug.
 

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