Posted by Phil on Saturday, May 5, 2012 at 9:07pm.
Two block are connected by a rope that runs over a pulley. The block on the tables has mass 4kg, the hanging block has mass 2kg, and the pulley has mass 0.5kg and radius 0.25m. Assume that the table is frictionless. If the block are released from the rest, determine their speeds after the hanging block has dropped 0.75m.
Please explain the formula and steps. Thanks.

Physics 121  Elena, Sunday, May 6, 2012 at 12:34pm
m1 =4 kg, m2 = 2 kg, m = 0.5 kg, R = 0.25 m, h= 0.75 m.
Projections of the equation according to the 2 Newton's law for two blocks on the horizontal (for the 1st block)and on the vertical (fot the 2nd block) axis:
m1•a = T1
m2•a =m2•gT2,
The equation of the pulley motion (2nd Nerton's law, for the rotational motion)
I•ε =M.
The moment of inertia of the pulley (disk) is
I =m•R²/2 ,
M = torque = (T1T2)•R,
ε = a/R,
I•ε =M => m•R²•a/2•R =(T1T2) •R =>
m•a/2 = (T1T2).
m1•a + m2•a = T1 + m2•g T2 = m2•g + (T1T2) = m2•g +m•a/2,
a = m2•a/[m1+m2m(m/2)] =
= 2•9.8/(4+2+0.125)=3.336 m/s^2,
a = v^2/2•h ,
v=sqrt(2•a•h) = sqrt(2•3.336•0.75) =
= 2.2 m/s^2
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