Pulley with Mass: Impact on Energy and Work Questions - Explained

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In energy and work calculations involving pulleys, the mass of the pulley should be considered, as it contributes to the system's total kinetic energy. A pulley with mass will have its own velocity and acceleration, resulting in additional kinetic energy that must be accounted for in the overall work done. If a system starts from rest, the total work done is equal to the kinetic energy of both the system and the pulley, implying that the system will have less kinetic energy for the same work output. The discussion raises questions about the impact of friction on the pulley’s rotation and energy transfer, suggesting that friction could influence the system's efficiency. Overall, the presence of a massive pulley alters the dynamics of the system, affecting both distance and velocity.
devanlevin
in all energy and work questions i have had till now, the mass of pulleys have not been taken into account, if they had, would the system's kinetic energy have increased?
it seems to me that a pulley with mass would have velocity and acceleration and therefore would have its own kinetic energy,
if the system started from resting point, then the total work would be W=EK1, ie kinetic energy of the system, whereas with a pulley with mass W=EK1+EK2 ie knetic energy of the system as it was and kinetic energy of the pulley, the system as a whole will manage the same work with less EK1, is this logical?? what does it mean? will it pass the same distance but with lesser velocity?
is this point only valid if there is friction between the pulley and the rope or would this still affect the rest of the system if there was no friction whatsoever?
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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