Find acceleration of a pulley system

[cantdisclosename]

Homework Statement

A pulley of radius, R and moment of inertia, I=2MR^2 is mounted on an axle with negligible friction. Block A, with a mass M, and Block B, with a mass of 3M, are attached to a light string that passes over the pulley. Assuming that the string doesn't slip on the pulley, Answer the following questions in terms of M, R, and fundamental constants.
a. What is the acceleration of the two blocks?
b. What is the tension force in the left section of the string? (M is on the left hanging)
c. What is the tension force in the right section of the string? (3M is on the right)

Homework Equations

Fnet=ma
τ=Iα

The Attempt at a Solution

a. Fnet=ma
3Mg-Mg=4Ma
a=g/2
Answer: g/3

For b and c. even if I plug in the right acceleration I get the wrong answer for b but right answer for c
b. Fnet=ma
Mg-T=Ma
Mg-T=Mg/3
-T=Mg/3-3Mg/3
-T=-2/3Mg
T=2/3Mg
Answer= 4/3Mg

c. Fnet=ma
3Mg-T=ma
3Mg-T=3M(g/3)
3Mg-T=Mg
T=2mg
Answer: 2mg

 

[John Morrell]

first of all, the total inertia of the system is going to include both blocks and the axle. If the blocks accelerate with a magnitude of a, then the pulley will experience an angular acceleration of alpha=a/R, since the acceleration of the outer-most part of the pulley is alpha*R. When you find the acceleration of any of the blocks, you have to take this into account.

For part c. you have to keep straight which direction you have defined as positive. If you have decided that the acceleration is positive, than whatever direction that acceleration is in must also be positive. In this case, the force of gravity will be negative and the tension will be positive. You have that mixed up, I believe.

 

[cantdisclosename]

first of all, the total inertia of the system is going to include both blocks and the axle. If the blocks accelerate with a magnitude of a, then the pulley will experience an angular acceleration of alpha=a/R, since the acceleration of the outer-most part of the pulley is alpha*R. When you find the acceleration of any of the blocks, you have to take this into account.

For part c. you have to keep straight which direction you have defined as positive. If you have decided that the acceleration is positive, than whatever direction that acceleration is in must also be positive. In this case, the force of gravity will be negative and the tension will be positive. You have that mixed up, I believe.

τ=Iα
T1R-T2R=(2MR^2)(a/R)
T1-T2=2Ma
a=(T1-T2)/2M
Fnet=ma
3Mg-Mg=4Ma
2Mg=(4M)((T1-T2)/2M)
and it leads me nowhere. I have no idea what to do because now a is not even a variable in the equation

 

[TomHart]

T1R-T2R=(2MR^2)(a/R)

This equation looks good.

3Mg - Mg = 4Ma
This equation not so good.
(Edit: Sorry, I originally referenced the wrong equation.)

I recommend that you draw a free body diagram for block A and block B and write equations relating sum of forces and acceleration for both of those (Edit: I should say each of those).
With those 2 equations and your first sum of torques equation, you should end up with three equations and three unknowns (T1, T2, and a).

And welcome to Physics Forums.

 

[cantdisclosename]

This equation looks good.

3Mg - Mg = 4Ma
This equation not so good.
(Edit: Sorry, I originally referenced the wrong equation.)

I recommend that you draw a free body diagram for block A and block B and write equations relating sum of forces and acceleration for both of those (Edit: I should say each of those).
With those 2 equations and your first sum of torques equation, you should end up with three equations and three unknowns (T1, T2, and a).

And welcome to Physics Forums.

I came up with the extra two equations I think they are:
1) 3Mg-T1=3Ma
2) Mg-T2=Ma
I tried using T1-T2=2Ma (equation 3) along with the two equations I made but still can't figure out acceleration.
I tried solving equations 1 and 2 for a and set them equal to each other but it ends up being T1=T2 and if I try substituting equation 3 into T1=T2 I end up with nothing again. I can't figure out what I'm missing.
And thank youEDIT: also tried using substituting equation 3 into both equations 1 and 2 but still nothing because I end up with 2 variables

 

[TomHart]

1) 3Mg-T1=3Ma
2) Mg-T2=Ma

Your torque equation has 3 variables: T1, T2 and a.
Solve your number 1 equation for T1 and plug the result of that into your torque equation.
Solve your number 2 equation for T2 and plug the result of that into your torque equation.

Doing that will leave you with only one variable (a) in your torque equation.

But also, I think you have a sign error in either equation 1 or equation 2. Make sure you are using the same sign convention as you used in your torque equation.

 

[cantdisclosename]

Your torque equation has 3 variables: T1, T2 and a.
Solve your number 1 equation for T1 and plug the result of that into your torque equation.
Solve your number 2 equation for T2 and plug the result of that into your torque equation.

Doing that will leave you with only one variable (a) in your torque equation.

But also, I think you have a sign error in either equation 1 or equation 2. Make sure you are using the same sign convention as you used in your torque equation.

Thank you so much! I figured it out