How Does Torque Affect Tensions in an Atwood's Machine with a Massive Pulley?

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In an Atwood's machine with a massive pulley, the tension T2 must be greater than T1 due to the need for the pulley to accelerate. The system's acceleration can be determined by analyzing the forces acting on both masses and the pulley. The tensions T1 and T2 can be calculated using Newton's second law and the rotational dynamics of the pulley. The pulley acts as a third body, requiring free body diagrams (FBDs) for accurate analysis. Understanding these dynamics is crucial for solving the problem effectively.
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An Atwood's machine consists of blocks of masses m1 = 8.4 kg and m2 = 20.7 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 7.40 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

(a) Why must the tension T2 be greater than the tension T1?

(b) What is the acceleration of the system, assuming the pulley axis is frictionless?

(c) Find the tensions T1 and T2.

Does anyone have idea how pully w/ mass works?
 
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Treat the pulley as a third body and draw three FBDs. Since the cord does not slip on the pulley, you have the tensions acting on the two sides of the pulley acting to give it an angular acceleration.
 
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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