Sorry for making an entire thread for this, but I'm completely stuck and didn't think this fit in the Math help thread.
The question asks: Given a 4.0 kg block 5.0 m above the ground and a 2.0 kg block on the ground connected by a rope hanging from a pulley with radius 0.16 m and moment of inertia 0.56 (kg*m^2), find the speed of the 4.0 kg block the moment it hits the ground using energy methods
What I've got so far:
PE
i = PE
f + KE + RE
m
1gh = m
2gh + (1/2)m
1v
2 + (1/2)m
2v
2 + (1/2)I(omega
2)
4*5*9.81 = (2*5*9.81) + ((1/2)*(4)*(v^2) + (1/2)*(2)*(v^2)) + ((1/2)*I*(omega^2)) (v^2 is the same for both blocks)
10g = (3v2) + ((1/2)*I*(omega^2))The problem I'm having is with the angular velocity omega. Using the equations:
omega
2 = 0
2 + 2 * angular acceleration * delta theta
delta theta = 5 / r
angular acceleration = ||torque|| / I
||torque|| = ||<-r,0,0> x <0,-4g,0> + <-r,0,0> x <0,-2g,0>|| = 2rg
I get:
omega
2 = 2*(2rg/I)*(5/r) = 20g/I
Problem is, plugging that value back into the conservation of energy equation gives:
10g = (3v
2) + (1/2)*I*(20g/I), or
10g = (3v
2) + 10g
which doesnt work out.
I've found the answer in the back of the book, and it's 2.65 m/s, but I have no clue how they got this answer.
By converting angular velocity to tangential velocity (since the rope doesn't slip), I got 2.99, which is pretty close, but still wrong, and I still have no clue where to go.
Could anyone help me out? Thanks for your time.