Question

Do the inertias of the weights of an Atwood's Machine affect the pulley's angular acceleration?

Answer 1

no, the idealised pully is massless.

Answer 2

That's irrelevant, but let's assume the pulley is not massless.

Answer 3

i interpreted your question in the only way that could make any sense in the physical world and i get this terse response?! C'mon.

OF COURSE the masses affect its angular acceleration. the assumption is that it has no inertia itself and that this is a straight mechanical pass through. however, it is connected to the masses and in the bizarre event you thought its angular acceleration was somehow relevant to anything, you could calculate it.

of course, the radius of the machine is also completely irrelevant in the basic Atwood calculation.

Answer 4

as is it's mass, even though you say "let's assume the pulley is not massless" in which case you are no longer describing an Atwood machine

Answer 5

Eh, an Atwood's Machine can certainly have a massive pulley. It's only when we idealize it for simplicity's sake that the pulley becomes massless. What I'm trying to figure out is how the inertias of the weights affect the angular acceleration of a massive pulley. From what I've been reading online, they don't, but I don't understand why. For example, if the machine's pulley had a moment of inertia of 1kg/m^2, and the weights were 1 and 2 kg, the pulley would supposedly accelerate at the same rate as a machine with the same 1kg/m^2 pulley, but with weights which weighed 1,001 and 1,002 kg. From what I understand, the difference in the weights is all that matters, but I can't figure out why the inertia of the weights doesn't matter.

Answer 6

if you work through the basic force equations, you will get something like this:

\[a = \frac{m2 - m1}{ \frac{M}{2} + m1 + m2}g\]

M is the pulley mass, and the only real assumption there is that its angular inertia is (1/2) M R^2, ie the pulley is a uniform disc or cylinder.

Answer 7

1/2 MR^2 is exactly the formula for the moment of inertia for a cylinder, so that was expected.

Answer 8

Basically, the basic force equations are what I'm asking about. They don't include the inertia of the masses; only the torques they impart unto the pulley matter in the calculations. However, they are moving with the system, so it seems as if their inertia should come into play, but it doesn't, and I'm curious why.

Answer 9

But wait! Maybe they do include their inertias in a way. The basic force equation is masses subtracted over masses added. if the masses were 1 and 2, the angular acceleration would be much larger than if the masses were 1001 and 1002. So larger masses would result in a commensurately smaller angular acceleration, which is what we would expect if the inertias were taken into account.

Answer 10

the equation says it all.

start with your example: masses of 1 & 2kg vs masses of 1001kg & 1002 kg.

the numerator is 1•g in both, but the denominator is very different.

in fact the "inertia" of all 3 go into the denominator and .....

Answer 11

Wait, only the ratio matters. If the masses are 5 and 10 or 500 and 1000, the acceleration will be the same.

Answer 12

So back to square one.

Answer 13

Starting equations should be
\[m_1a=m_1g-T_1\]
\[m_2a=T_2-m_2g\]
\[\frac{MR^2}{2}(\frac{a}{R})=(T_1-T_2)R\]

Answer 14

which gives @IrishBoy123 's result. here I am assuming m1 is heavier than m2