Angular momentum/ energy conservation.

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SUMMARY

The discussion revolves around the calculation of energy dissipation in a system involving a block and a pulley. The block, with a mass of 0.7 kg, descends a distance of 0.79 m, achieving a speed of 1.233 m/s. The moment of inertia of the pulley is given as 0.0057 kg*m². The equation used for energy conservation, mgh = 1/2 mv² + mg(h final) + 1/2 iw² + L (energy loss), is questioned for its correctness in this context, indicating potential miscalculations or missing variables.

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  • Understanding of classical mechanics principles, specifically energy conservation.
  • Familiarity with rotational dynamics, including moment of inertia and angular velocity.
  • Knowledge of kinematic equations for motion under gravity.
  • Ability to manipulate algebraic equations involving multiple variables.
NEXT STEPS
  • Review the principles of energy conservation in rotational systems.
  • Learn about the relationship between linear and angular motion, particularly the equations involving moment of inertia.
  • Investigate the effects of friction on energy dissipation in mechanical systems.
  • Practice solving similar problems involving pulleys and blocks to reinforce understanding of energy loss calculations.
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Roland of G
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Homework Statement


A block of mass 7.00×10-1kg is suspended by a string which is wrapped so that it is at a radius of 6.00×10-2m from the center of a pulley. The moment of inertia of the pulley is 5.70×10-3kg*m2. There is friction as the pulley turns. The block starts from rest, and its speed after it has traveled downwards a distance of D=0.790m is 1.233m/s. Calculate the amount of energy dissipated up to that point.
I set up an equation like this:
mgh=1/2mv^2+mg(h final)+1/2iw^2+L(energy loss) This makes sense to me... Why does this not work correctly?


Homework Equations





The Attempt at a Solution


mgh=1/2mv^2+mg(h final)+1/2iw^2+L(energy loss) This makes sense to me... Why does this not work correctly?
 
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Welcome to PF!

Hi Roland of G! Welcome to PF! :smile:

(have an omega: ω and try using the X2 icon just above the Reply box :wink:)
Roland of G said:
mgh=1/2mv^2+mg(h final)+1/2iw^2+L(energy loss) This makes sense to me... Why does this not work correctly?

It should work. :confused:

Show us the figures. :smile:​
 

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