Find acceleration of a pulley system

• cantdisclosename
In summary: Mg-T1=3Ma2) Mg-T2=MaYour 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.In summary, the conversation involved solving a physics problem involving a pulley with
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)

Fnet=ma
τ=Iα

The Attempt at a Solution

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

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

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

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.

John Morrell said:
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

cantdisclosename said:
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.

TomHart said:
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

cantdisclosename said:
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.

TomHart said:
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

TomHart

1. How do you find the acceleration of a pulley system?

The acceleration of a pulley system can be found using the following formula: a = (m1 - m2)g / (m1 + m2), where m1 is the mass of the heavier object and m2 is the mass of the lighter object. g represents the gravitational acceleration, which is typically 9.8 m/s^2. This formula assumes that the pulleys and ropes are massless and there is no friction.

2. What factors affect the acceleration of a pulley system?

The acceleration of a pulley system is affected by several factors, including the mass of the objects attached to the pulleys, the number of pulleys in the system, and the angle at which the ropes are pulled. Additionally, friction and the mass of the pulleys themselves can also affect the acceleration.

3. How does adding more pulleys affect the acceleration of a pulley system?

Adding more pulleys to a system can actually decrease the acceleration. This is because each additional pulley adds more friction to the system, as well as more mass. However, using more pulleys can also allow for heavier objects to be lifted, as the weight is distributed among multiple pulleys.

4. Can the acceleration of a pulley system be negative?

Yes, the acceleration of a pulley system can be negative. This would occur if the heavier object is on the lighter side and the lighter object is on the heavier side, causing the system to move in the opposite direction of the gravitational acceleration. Negative acceleration can also occur if there is friction in the system, which would slow down the overall acceleration.

5. How does the angle of the ropes affect the acceleration of a pulley system?

The angle of the ropes can have a significant impact on the acceleration of a pulley system. The greater the angle, the more tension is required to lift the objects, which in turn increases the acceleration. However, if the angle is too large, it can also lead to more friction and decrease the acceleration. The ideal angle for maximum acceleration is typically between 30-45 degrees.

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