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.

Question

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!

Answer 1

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.

Answer 2

The acceleration should come out as 0.1g actually.
So, v = sqrt(2*a*s) = sqrt(2*0.1g*1) = 1.40m/s

Answer 3

get outta here uiuc boii