Algebraic Manipulation of Equations

1. Nov 17, 2014

EzequielSeattle

1. The problem statement, all variables and given/known data
I have two equations. The first is for all of the forces on a hanging mass from a pulley. The second is for the sum of the torques about the pulley from which the mass hangs. I simply have to combine the equations to find the acceleration of the object. I have attempted every algebraic manipulation I can think of and keep coming out with the wrong answer. Please help.

2. Relevant equations
T-mg=ma (for the sum of forces on the hanging mass)
Tr=I(-a/r) (for the torques about the pulley)

Here, T is tension, m is the mass of the hanging object, a is the acceleration, r is the radius of the pulley, I is the moment of inertia of the pulley.

I'm supposed to combine the two equations to eliminate T and solve for a.

3. The attempt at a solution
OK, solve equation 1 for T.

T = ma + mg

Cool, now plug into equation 2.

(ma+mg)r=I(-a/r)
mr(a+g)=I(-a/r)
a+g=I(-a/mr^2)
1+g/a=I/mr^2
g/a=-I/mr^2-1
a = -(gmr^2)/(I)-g

I keep coming out with the same exact solution every time, but it is apparently wrong. Can someone tell me where I went wrong?

2. Nov 17, 2014

Simon Bridge

a+g=I(-a/mr^2)
1+g/a=I/mr^2

I think you mislaid a minus sign there.
But you are also going about it the long way.

(ma+mg)r=I(-a/r)

That simplifies to $mr^2a+mr^2g = -Ia$
...now get all terms involving "a" on the LHS and put everything else of the RHS.

3. Nov 17, 2014

EzequielSeattle

a(mr2 I) = -mgr2

a = -(mgr2)/(mr2+I)

Thank you, that's correct. I could cry tears of joy.

4. Nov 17, 2014

Simon Bridge

No worries - for the future, it is often useful to try getting rid of all the denominators so you can write the equation out on one line.
Makes the equations easier to type too.
After that it's just a matter of grouping the term you want to solve for on one side.