How Does Frictionless Pulley Affect Acceleration in a Modified Atwood's Machine?

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SUMMARY

The discussion focuses on calculating the acceleration of a modified Atwood's machine featuring a frictionless pulley. The system consists of a 3 kg mass on a table and a 4 kg hanging mass, with the pulley having a moment of inertia of 1/2 kg*m² and a radius of 0.3 m. Key equations utilized include F = Ma, T = Ia (angular), and the relationship between torque and tension. The critical insight is that the tensions on either side of the pulley differ due to its moment of inertia, necessitating the use of two distinct tension variables, T1 and T2, to solve the system of equations.

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Homework Statement



Given the system shown, if the pulley has frictionless bearings, find
a, the magnitude of the acceleration of the masses.

This is a modified Atwood's machine, instead of the 2nd mass hanging however, it is stationary on a table. The pulley is at the edge of the table and the 1st mass is hanging down. The mass on the table has a mass of 3kg, the hanging mass is 4kg. The pulley has an I of 1/2 kg*m^2 and a radius of .3m.

Homework Equations



F = Ma
T = Ia(angular)
T = r x F

The Attempt at a Solution



F = Ma
Mg - T2 = ma
also Torque = r x F
Torque = rTsin90
RT = Ia(angular)
rT = I (A/R) --> because linear accel = angular * radius

But from here I'm lost...
 
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Vixenbergen said:
Given the system shown
? There is no system shown. Did you forget your diagram?
 
The critical point here is that the tension of the rope is not equal between the two sides of the pulley because the pulley has a moment of inertia hence it needs torque to rotate, and the torque would be zero if the two tensions were equal. So we have to use ##T_1,T_2## for the two tensions, and the system of equations is (by index 1 I mean the mass that is hanging)
$$m_1g-T_1=m_1a$$
$$T_2=m_2a$$
$$(T_1-T_2)R=I\frac{a}{R}$$

Linear system of three equations with three unknowns, shouldn't be much of a trouble to solve for a college student.
 
phinds said:
? There is no system shown. Did you forget your diagram?
I think that here we have what is more widely known as a "half-Atwood" machine.
 

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