Posted by **Jake** on Wednesday, April 18, 2012 at 10:15am.

Two masses, m1=3kg and m2=2kg, are suspended with a massless rope over a pulley of mass M = 10kg. The pulley turns without friction and may be modeled as a uniform disk of radius R=.1m. You may neglect the size of the masses. The rope does not slip on the pulley. The system begins at rest.

If m1 starts at a height h=1m above the ground, what is its speed v when it hits the ground?

I've tried a few different ways to get it but I can't seem to get the answer. I've tried using the sum of the torques as well as trying to use energy. I can't seem to figure it out, any help is much appreciated!

- Physics -
**Elena**, Wednesday, April 18, 2012 at 5:47pm
In vector form

m1•¬a = m1•¬g - ¬T1

m2•¬a = m2•¬g - ¬T2

I•¬ε =¬M

Projections on the vertical axis:

m1•a = m1•g - T1,

m2•a = m2•g – T2,

(mR^2/2)•(a/R) = (T1-T2)•R.

Solving this system for acceleration a, we obtain

a= (m1-m2)•g/(m1 +m2+m/2) =(3-2)/(3+2+10/2) = 0.1 m/s^2.

Since

s =v^2/2•a.

v =sqrt(2•a•s) = sqrt(2•0.1•1)= 0.45 m/s.

- Physics -
**Jose**, Wednesday, December 16, 2015 at 3:54pm
The acceleration should come out as 0.1g actually.

So, v = sqrt(2*a*s) = sqrt(2*0.1g*1) = 1.40m/s

- Physics -
**fook you m8**, Wednesday, April 20, 2016 at 3:21pm
get outta here uiuc boii

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