Solving Atwood's Pulley Homework Equations

[mlostrac]

Homework Statement

Equation 1: m1g - T1 = m1a (for first mass)
Equation 2: T2 - m2g = m2a (for second mass)
Equation 3: (T1-T2)R = IA (for pulley; A = rotational acceleration)

What are the steps I must take to get the final equation that describes the motion of the system:
(m1-m2) g = (m1+m2+I/R^2)a ??

Thank you

Homework Equations

The Attempt at a Solution

I can figure out a = g(m1-m2)/(m1+m2) BUT i don't know how to factor in equation 3?

 

[pgardn]

Homework Statement

Equation 1: m1g - T1 = m1a (for first mass)
Equation 2: T2 - m2g = m2a (for second mass)
Equation 3: (T1-T2)R = IA (for pulley; A = rotational acceleration)

What are the steps I must take to get the final equation that describes the motion of the system:
(m1-m2) g = (m1+m2+I/R^2)a ??

Thank you

Homework Equations

The Attempt at a Solution

I can figure out a = g(m1-m2)/(m1+m2) BUT i don't know how to factor in equation 3?

Are you trying to find the linear acceleration of the system?

 

[Doc Al]

I can figure out a = g(m1-m2)/(m1+m2)

That's only true for a massless pulley.

BUT i don't know how to factor in equation 3?

Hint: Express the rotational acceleration A in terms of a. Then you'd have 3 equations and 3 unknowns.

 

[mlostrac]

Are you trying to find the linear acceleration of the system?

No, I'm trying to show how to derive the last equation from the first 3.

Doc Al:
I know that A =a/R, which would give me (T1-T2)R = I (a/R) , I'm just not sure how to combine that with the two equations I combined for the masses: a = g(m1-m2)/(m1+m2)

 

[Doc Al]

Doc Al:
I know that A =a/R, which would give me (T1-T2)R = I (a/R) ,

Good. That becomes your revised third equation.

I'm just not sure how to combine that with the two equations I combined for the masses: a = g(m1-m2)/(m1+m2)

Careful: You did not get this equation by combining the first two equations! The first two equations have three unknowns--a, T1, T2--there's no way to eliminate the two tensions using just the first two equations. (You would get this equation if you combined the two equations for an Atwood's machine with a massless pulley. But those equations are different, since there is only a single tension to worry about.)

To combine the first two equations, you just add them. To combine all three, just add all three!

 

[mlostrac]

Good. That becomes your revised third equation.

Careful: You did not get this equation by combining the first two equations! The first two equations have three unknowns--a, T1, T2--there's no way to eliminate the two tensions using just the first two equations. (You would get this equation if you combined the two equations for an Atwood's machine with a massless pulley. But those equations are different, since there is only a single tension to worry about.)

To combine the first two equations, you just add them. To combine all three, just add all three!

Ok, so I was wrong by assuming a massless pulley therefore making that first equation I solved for "a" wrong.

Now I have:
Equation 1: m1g - T1 = m1a (for first mass)
Equation 2: T2 - m2g = m2a (for second mass)
Equation 3: (T1-T2)R = I(a/R) (for pulley)

Do I set 1 and 2 equal to zero, and then make them equal to each other so I can isolate T1 and T2 to make (T1-T2). Then do I sub (T1-T2) into equation 3?

Im confused

 

[Doc Al]

Do I set 1 and 2 equal to zero, and then make them equal to each other so I can isolate T1 and T2 to make (T1-T2).

I don't really know what you mean by setting them equal to each other or setting them equal to zero.

In any case, it's much easier than all that.

Why don't you do what I suggested in the last sentence of my previous post?

 

[mlostrac]

Ok I think I am starting to get it, I got this far:

Adding equations 1 and 2: m1g + (-m2g) - T1 + T2 = m1a + m2a
Adding in equation 3: g (m1-m2)+ (T1-T2) = a (m1+m2) + I (a/R^2)

g(m1-m2) = a (m1+m2+I/R^2)

Everything look right?

 

[Doc Al]

Ok I think I am starting to get it, I got this far:

Adding equations 1 and 2: m1g + (-m2g) - T1 + T2 = m1a + m2a

Good.

Adding in equation 3: g (m1-m2)+ (T1-T2) = a (m1+m2) + I (a/R^2)

I forgot to mention that you needed to divide both sides of equation 3 by R to give units of force. So equation 3 becomes: (T1-T2) = I(a/R^2)

Adding that, you get:
g (m1-m2) - T1 +T2 + (T1-T2) = a (m1+m2) + I (a/R^2)

Which gives you your final result:

g(m1-m2) = a (m1+m2+I/R^2)

Everything look right?

Looks good!

 

[mlostrac]

Good.

I forgot to mention that you needed to divide both sides of equation 3 by R to give units of force. So equation 3 becomes: (T1-T2) = I(a/R^2)

Adding that, you get:
g (m1-m2) - T1 +T2 + (T1-T2) = a (m1+m2) + I (a/R^2)

Which gives you your final result:

Looks good!

Thanks so much for your help Doc Al! Very appreciated