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

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Homework Help Overview

The discussion revolves around a modified Atwood's machine where one mass is stationary on a table and the other mass is hanging. The pulley is described as having frictionless bearings, and participants are exploring how this affects the acceleration of the masses involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion and the role of tension in the system. Some question the absence of a diagram, while others highlight the need to consider different tensions on either side of the pulley due to its moment of inertia.

Discussion Status

The conversation is ongoing, with some participants providing insights into the equations that govern the system. There is recognition of the complexity introduced by the pulley’s moment of inertia, and multiple interpretations of the problem setup are being explored.

Contextual Notes

There is a noted absence of a diagram, which some participants indicate is critical for understanding the system. The discussion also references the specific characteristics of the modified Atwood's machine, which may influence the analysis.

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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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