- #1
vetgirl1990
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Homework Statement
A block of mass 2kg (mb) can slide down a frictionless 53 degree inclune, but it is connected to a pulley of mass 4kg (mp with a radius of 0.5m. The pulley may be treated as a disk.
What is the angular acceleration of the pulley?
Homework Equations
a = rα
The Attempt at a Solution
I first expressed each object in terms of Newton's 2nd Law
BLOCK:
ΣFx = mbsinθ - T = mba
ΣFy = N - mgsinθ = 0
PULLEY:
ΣFx = T = mmpa
ΣFy = N-mg = 0
Then, isolated for T (tension) for both the block and pulley and equated them to each other. Their tension would be the same because the pulley and block are connected by the same rope.
(1) T = mbgsinθ - mba
(2) T =mpa
mpa = mbgsinθ - mba
a = mbgsinθ / (mb + mp)
a = 2.6m/s
α = a/r = (2.6m/s/s) / 0.5m = 5.22rads/s/s
CONCERNS:
We weren't given the final answer, so I'm not sure if my approach is correct. But I have a feeling that my Newton's 2nd Law (N2L) expression for the pulley is incorrect.
ΣFx = T = mpa
Since the pulley is rotating, would the acceleration in N2L be CENTRIPETAL ACCELERATION?
However, if this was the case, then the expression would turn into: ΣFx = T = mpv2/r
This introduces another unknown -- velocity! If this were the case, I don't know how to solve the problem.
Thank you!