Algebraic Manipulation of Equations

[EzequielSeattle]

Homework Statement

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.

Homework 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.

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?

 

[Simon Bridge]

About line 3 and 4...

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.
When you had:

(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.

 

[EzequielSeattle]

a(mr² I) = -mgr²

a = -(mgr²)/(mr²+I)

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

 

[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.

 

[HalfLostAndSearching]

It seems like you have made a mistake in the last step where you have divided by a on both sides. This is incorrect as a is not a constant in the equation and therefore cannot be divided out. Instead, you should keep a on one side and move all the other terms to the other side. This will result in the correct solution of:

a = -(gmr^2)/(I + mr^2)

This shows that the acceleration of the object is dependent on the mass, radius of the pulley, and moment of inertia of the pulley. It is important to double check your algebraic manipulations and make sure you are following the correct order of operations. Additionally, in scientific work, it is always good practice to double check your calculations and use units to make sure your final answer makes sense.