Arturo O. answered • 11/28/17
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Becky T.
And those are the correct numbers in the problem.
Now, I have seen τ=Iα, and I gather that mgr = Iα comes from τ=r*F, where F is equal to §ma, but i have not seen acceleration= to radius*angular acceleration. Is that a common equation or is that another derivation (manipulation of another equation)?
§Also, F turns into mg (instead of staying ma) because the object is falling, correct?
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11/28/17
Arturo O.
The torque on the wheel is due to the weight transmitted by the cord to the radius of the wheel, so
τ = Fr = (mg)r
You know that
τ = Iα
It is also a common equation that the tangential acceleration of a point located a distance r from the rotation axis is
a = rα
So the equation
a = mgr2/I
is correct. Plug in the given numbers and get 4.88 m/s2.
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11/28/17
Becky T.
I see, & you can think of no reason why 3.26 m/s^2 might be the answer?
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11/28/17
Arturo O.
The torque on the wheel is due to the tension T in the cord applied at the radius of the wheel.
τ = Tr = Iα = Ia/r
For the weight, with positive being up, and downward acceleration,
T - mg = m(-a) ⇒ T = m(g - a)
Combine these 2 equations and get
m(g - a)r = Ia/r
Solve for a and get
a = mg / (I/r2 + m) = (3.63)(9.8) / [2.71/(0.610)2 + 3.63] m/s2 = 3.26 m/s2
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11/28/17
Arturo O.
See a revised solution in my last Comment below. I did not read the original problem statement carefully! I found and fixed my error. Ignore my previous Answer and previous Comments. Let me know if you have any questions.
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11/28/17
Arturo O.
I meant to say "in my last Comment above." You can see that 3.26 m/s2 is correct, per the revised solution in my Comment above.
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11/28/17
Becky T.
11/28/17