Torque and Tangential Acceleration

welterweight08

1. The problem statement, all variables and given/known data

My ENTIRE AP physics class is stumped on the way to correctly write the formula to find acceleration for a falling block (mass of "m") attached by a string (tension of "t") to a fixed, rotating pulley (mass of "M", radius of "r"). Our teacher told us that for a block attached on the left side of a pulley (rotating counterclockwise) the tangential acceleration ("a") of the pulley should be positive (+) and for one on the right side (pulley rotating clockwise) a should be negative (-). We cannot come to a decisive conclusion as to what the final equation to find a for each side should be.

m = block mass
M = pulley mass
r = radius of pulley
t = tension of string
a = tangential acceleration (acceleration of the block)
I = Inertia of pulley
A = angular acceleration of pulley
g = Gravity (assume g=10 if needed)

2. Relevant equations

Positive "a" Formula Set:



(+)Formula One: Pulley Rotating CCW


T = rt, T = IA , I = 1/2M(r^2), and A = a/r

therefore... rt = 1/2 M(r^2)(a/r)
(r^2)t = 1/2M(r^2)a
t = 1/2Ma

(+)Formula Two: Block Falling to the Left

t = sum of all forces
t = mg + ma

Negative "a" Formula Set:



(-)Formula One: Pulley Rotating CW

T = rt, T = IA , I = 1/2M(r^2), and A = -a/r

therefore... rt = 1/2 M(r^2)(-a/r)
(r^2)t = 1/2M(r^2)(-a)
t = -(1/2Ma)

(-)Formula Two: Block Falling to the Right

t = sum of all forces
t = mg + [m(-a)]
or
t = mg - ma
(This part is where we some of the confusion begins. We assume t = mg "-" ma since the block is falling to the right and "a" should be negative)


3. The attempt at a solution

Positive, CCW, Left

t = mg + ma
and
t = 1/2Ma
so
1/2Ma = mg + ma
both sides x 2
Ma = 2mg + 2ma
both sides -2ma
Ma - 2ma = 2mg
factor out a
a(M - 2m) = 2mg
both sides / (M-2m)
a = 2mg / (M - 2m)

Negative, CW, Right

t= mg - ma
and
t = -(1/2Ma)
so
-(1/2Ma) = mg - ma
both sides x-(2)
Ma = -2mg + 2ma
both sides -2ma
Ma - 2ma = -2mg
actor out a
a(M - 2m) = -2mg
both sides / (M - 2m)
a = -2mg / (M - 2m)

I've worked these out to the best of my ability and I believe that they are correct. Can someone please either confirm this or correct me ASAP?
Thanks!

AlephZero

In both cases, you are taking t to be positive when there is a tension in the string. For the lower block, the correct equation is always t = mg + ma if you measure a as positive downwards.

It is obvious physically that the lower block moves downwards, therefore the acceleration a, measured positive downwards, must be positive.

Your "negative, CW, right" answer is wrong. The mistake is saying t = mg - ma.

welterweight08

Thanks, I'll explain this to the rest of my class!