Why Does the Atwood Machine Accelerate Faster with Increased Mass on One Side?

AI Thread Summary
The Atwood machine accelerates faster with increased mass on one side due to the greater net force resulting from the difference in gravitational force acting on the two masses. As the mass on one side increases, it creates a larger downward force, leading to greater acceleration according to Newton's second law. The inertia of the system also plays a role, as more mass increases the system's overall inertia, affecting the torque on the wheel. Calculating the forces involved can demonstrate how the acceleration changes with varying mass. Understanding these principles clarifies the mechanics behind the Atwood machine's operation.
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Homework Statement


Well gravity's acceleration on objects is 9.8 m/s^2 and the question is why the Atwood Wheel moves faster as the mass on side is increased.


Homework Equations


N/A


The Attempt at a Solution


I am thinking it has something to do with inertia and the torque of the wheel since the more mass more inertia if i am not mistaken. Any help or point in the right direction would help thank you
 
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you could just simply do some proofs by just proving the masses have a higher force by simply crunching numbers like 5 times for each side one light on heavy and increase mass for the light one and increas for heavy and show how the force increase and drops. pretty simple but shou8ld prove it
 
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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