Homework Help: Aceleration of systems

1. Jun 20, 2012

jesusjones

I need some clarification here...

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

There are two masses and a pulley.

The pulley has R = .12m and I = .09 kgm^2

Mass 1 in hangs freely in the air and is 4kg

mass 2 is on a table(no friction) and is 6kg

I am finding the magnitude of the accel of mass 1

2. Relevant equations

3. The attempt at a solution

I set this up to solve for accel of m1

Ma = m1g - T - m2g (where the M is the mass of both blocks)

T(tension) = F(force) = τ/r(torque radius) = Ia/r^2 (inertia , accel of blocks radius)

and so Ma = m1g - Ia/r^2 - m2a

I end up with 1.7 m/s^2 but the answer should be 2.4 m/s^2.

Now if M instead of being the added mass of both blocks is changed to only the mass of m1 the answer is 2.4 m/s^2. But I am convinced that M should be the mass of both blocks. What is going on here?? why is it only the mass of block 1 or did I do something else wrong.
Or is possibly just the wording of the question. The accel of the system would be 1.7m?s^2 but the magnitude of the accel for m1 is 2.4 m/s^2 ( maybe??) I'm confused already and need this straightened out so I can get back on track

Thanks for your time,
Andrew

2. Jun 20, 2012

jesusjones

I seem to have figured this out for myself.

The Ma on the left side of my equation needed to be only the mass of the block in question. It seems that for systems of two blocks where one block is on a horizontal plane the M in Ma is only the block that is hanging freely and has accel due to gravity.

but if the plane is not horizontal then the M in Ma is the total mass of the system (m1 + m2)

Playing around with this http://hyperphysics.phy-astr.gsu.edu/hbase/incpl2.html#c1
and testing my formulas seems to show this is true. If anything I've said here is wrong please correct me.

3. Jun 21, 2012

MalachiK

I don't get this bit. What is the m2g term? The weight of the second block?

4. Jun 21, 2012

azizlwl

You have only one T.
So the net T is zero, no net torque to the pulley.

Like pulling 3 masses. The second/center should experienced net force(one pulling forward and other pulling back) for it to move.

Last edited: Jun 21, 2012
5. Jun 21, 2012

MalachiK

I wasn't sure about exactly how to set up the forces as I haven't done one of these for a while - but then I had the idea to use the conservation of energy, because I'm absolutely certain that this is true! Seems to me that...

m1gv = $\frac{d}{dt}$(1/2 m1v2)+$\frac{d}{dt}$(1/2 m2v2)+$\frac{d}{dt}$(1/2 (I/r2)v2)

I applied the product rule for differentiating each of the KE terms on the right and then fiddled with the algebra a bit before ending up with about 2.4 ms-2. So I'd say that the 2.4 ms-2 answer is correct - but you might not want to take my word for this!

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