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•g-T2,
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 = (T1-T2)•R,
ε = a/R,
I•ε =M => m•R²•a/2•R =(T1-T2) •R =>
m•a/2 = (T1-T2).
m1•a + m2•a = T1 + m2•g -T2 = m2•g + (T1-T2) = m2•g +m•a/2,
a = m2•a/[m1+m2-m(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
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 friction-less. 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.
1 answer
1 answer