Torque and Tangential Acceleration

AI Thread Summary
The discussion centers on deriving the correct formulas for tangential acceleration of a falling block attached to a rotating pulley. For a block on the left side of a counterclockwise pulley, the tension is expressed as t = mg + ma, while for a block on the right side of a clockwise pulley, the correct equation should reflect that the acceleration is positive downwards, leading to t = mg + ma as well. The initial confusion arose from incorrectly assuming t = mg - ma for the clockwise rotation, which was clarified as incorrect. The final equations for acceleration are a = 2mg / (M - 2m) for the counterclockwise case and a = -2mg / (M - 2m) for the clockwise case. Understanding the signs and direction of acceleration is crucial for solving these types of problems accurately.
welterweight08
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Homework Statement



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)

Homework Equations



Positive "a" Formula Set:

Left.jpg


(+)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:

Right.jpg


(-)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)

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