What does frictionless pulley really mean?

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A frictionless pulley refers to both the groove where the string moves and the axle, but neglecting the kinetic energy of the wheel allows for similar results regardless of friction in the axle. The key point is that a frictionless pulley does not dissipate energy, maintaining consistent force magnitude throughout the rope. This principle applies universally to ideal, frictionless pulleys, independent of their design specifics. The discussion clarifies that a frictionless axle with a massless wheel yields the same results as a frictionless groove in terms of energy equations. Understanding these mechanics is crucial for accurate analysis in physics.
andyrk
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When we say "pulley is frictionless", do we mean its groove where the string moves or its axle or both?
 
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The axle - but if you can neglect kinetic energy of the wheel, a frictionless groove gives the same result.
 
mfb said:
The axle - but if you can neglect kinetic energy of the wheel, a frictionless groove gives the same result.

How? How does it give the same result as an axle which has friction, if we neglect KE of the wheel?
 
The pulley will not dissipate any energy, and the magnitude of force is the same everywhere in the rope.
That is true for every ideal, frictionless pulley, it does not even depend on the details how you design the pulley.
 
andyrk said:
How? How does it give the same result as an axle which has friction, if we neglect KE of the wheel?
That's not what mfb wrote. mfb said that (frictionless axle with massless wheel) gives the same result (i.e. no role in the equations) as frictionless groove.
 
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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