Mass Pulley Acceleration Problem

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The discussion revolves around solving a physics problem involving two masses connected by a cord over a pulley. The key equations include torque, tension, and net force for both masses, with the correct acceleration of m1 being 3.13 m/s². Participants emphasize the need for a fourth equation relating angular acceleration to linear acceleration and highlight the importance of accurate signage in the equations. The consensus is to draw free body diagrams for clarity and to systematically solve the equations while plugging in the known values. Correcting the sign errors and completing the equations will lead to the solution.
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Homework Statement


A m2 = 4.63 kg mass is connected by a light cord to a m1 = 2.10 kg mass on a smooth surface (see the figure below).The pulley rotates about a frictionless axle and has a moment of inertia of 0.513 kg•m2 and a radius of 0.257 m. Assuming that the cord does not slip on the pulley, find the acceleration of m1.
15ecl94.jpg

Homework Equations


Torque = I*alpha = F*D

The Attempt at a Solution


FBD Mass 1
F_net = Tension 2... mass1*a = Tension 2
FBD Mass 2
F_net = mg-T1
T1 = mass2*g- mass2*a

Torque = I*alpha
Torque_1+Torque_2 = I*alpha
Tension_1+Tension_2 = I*alpha/r

and now I am stuck. Correct answer is 3.13 m/s^2
 
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Leeoku said:

Homework Statement


A m2 = 4.63 kg mass is connected by a light cord to a m1 = 2.10 kg mass on a smooth surface (see the figure below).The pulley rotates about a frictionless axle and has a moment of inertia of 0.513 kg•m2 and a radius of 0.257 m. Assuming that the cord does not slip on the pulley, find the acceleration of m1.
15ecl94.jpg



Homework Equations


Torque = I*alpha = F*D


The Attempt at a Solution


FBD Mass 1
F_net = Tension 2... mass1*a = Tension 2
FBD Mass 2
F_net = mg-T1
T1 = mass2*g- mass2*a

Torque = I*alpha
Torque_1+Torque_2 = I*alpha
Tension_1+Tension_2 = I*alpha/r

and now I am stuck. Correct answer is 3.13 m/s^2
You have noted 3 equations with 4 unknowns (alpha, a, T1 and T2).. You need a 4th that relates a and alpha. Also, watch your signage in your second and third equations.
 
Draw three free body diagrams; one for each mass and one for the pulley.
 
cant i just convert the alpha to a/r to get I*a/r^2. but then i don't get how to eliminate the massses.

Tension 2 takes the weight of both so... its (M1+M2)g
Tension one would be Tension 2 - mass2*g

im not sure if my logic is right
 
You just need to solve the system of equations.
 
Leeoku said:
cant i just convert the alpha to a/r to get I*a/r^2. but then i don't get how to eliminate the massses.

Tension 2 takes the weight of both so... its (M1+M2)g
Tension one would be Tension 2 - mass2*g

im not sure if my logic is right
No, it's not.

Now, get on with drawing the free body diagrams.
 
You had 3 of the 4 equations correct in your original post, except for a signage error, and you now have the 4th equation relating a and alpha. So correct your sign error, and, as jhae2.718 notes, solve the equations. Plug in the values of m1 and m2 before solving; they are given, so make life easier for you.
 
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