Rotational Motion: Energy Conservation

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SUMMARY

The discussion centers on solving a physics problem involving rotational motion and energy conservation. The scenario includes a sliding block with a mass of 0.850 kg, a counterweight of 0.420 kg, and a pulley with a mass of 0.350 kg. The solution utilizes the energy method, applying the equation ΔKE (initial) = ΔKE (final) to predict the block's velocity at a second photogate located 0.700 m away. The calculated final velocity is 1.5855 m/s, confirming the approach's validity.

PREREQUISITES
  • Understanding of kinetic energy (KE) and its equations
  • Familiarity with rotational inertia (I) and its calculation
  • Knowledge of friction coefficients and their impact on motion
  • Basic principles of Newton's laws of motion
NEXT STEPS
  • Study the derivation and application of the rotational inertia formula
  • Learn about energy conservation principles in mechanical systems
  • Explore the effects of friction on motion in physics problems
  • Investigate the relationship between linear and angular acceleration
USEFUL FOR

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

SpringWater
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Homework Statement


the sliding block has a mass of .850kg the counterweight has a mass of .420kg , and the pulley has a mass of .350kg with outer and inner radius .030m, .020m, the coefficient of friction is .250. the pulley turns the axle. the light cord does not stretch and does not slip on the pulley. The block has velocity of .820m/s towards the pulley when passes by photogate . use the energy method to predict the velocity it moved to a second photogate. .700m away?


Homework Equations





The Attempt at a Solution



I understand how to set up the problem using
ΔKE (initial)= ΔKE (final)

however i am wondering if this is possible...
Ke(m1)+KE(m2)+KE(rot)=(uk)(m1)(g)(X)

(V)^(2)*((.5)(m1)+(.5)(m2)+(.5)(I)(1/(r)^(2)))=(μk)(m1)(g)(X)

Where I=2.275E-4 & r=.02m & X=.7m By doing this i get V=1.5855 m/s

This is the correct answer! Did i get the correct answer just by coincidence or is this actually possible?
 
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Use the fact that W=ΔEk

So you get:

h=separation between photogates
a=acceleration of the system
m=mass of the block moving through the photogate
M=mass of the other block
v=velocity of the system

mah=0.5m(v22-v12)

So the only thing you need to know in order to solve for v2 in the equation above is a, which you can get using rotational formulas and Newton's laws.
 

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