pulley problem

[zekester]

a 4kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging 2kg block. the pulley is a uniform disk of radius 8cm and mass 0.6kg. find the speed of the 2kg block after it falls from rest a distance of 2.5m.

i just can't seem to figure it out.
Any help would be appreciated

[StephenPrivitera]

Use the work-energy theorem: Wnet=Kf-Ki. Since in your case the object is initially at rest, Ki=0. So Wnet=Kf=1/2mvf2. So vf=(2Wnet/m)1/2.
Also, since the forces acting on the body are mg and T are constant, Wnet=mgx-Tx. So,
vf=(2gx-2Tx/m)1/2
I've done most of the problem for you. I leave it to you to find T.

Also, since mg and T are constant, so is a, and you can use
vf=(2ax)1/2
What is a? a=g-T/m according to Newton's third law. So
vf=(2gx-2Tx/m)1/2, which agrees with what I showed already.

[formulajoe]

T = mg/2 correct?

[Doc Al]

Originally posted by zekester
a 4kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging 2kg block. the pulley is a uniform disk of radius 8cm and mass 0.6kg. find the speed of the 2kg block after it falls from rest a distance of 2.5m.

Use conservation of energy. Initially, all you have is PE (hint: measure PE from the final position), which is transformed to KE. Be sure to include the KE of all three bodies. (No need to find tensions or acceleration.)

[formulajoe]

i got 16.3 m/s as a final velocity.
is that even close?

[StephenPrivitera]

Originally posted by StephenPrivitera
Use the work-energy theorem: Wnet=Kf-Ki. Since in your case the object is initially at rest, Ki=0. So Wnet=Kf=1/2mvf2. So vf=(2Wnet/m)1/2.
Also, since the forces acting on the body are mg and T are constant, Wnet=mgx-Tx. So,
vf=(2gx-2Tx/m)1/2
I've done most of the problem for you. I leave it to you to find T.

Also, since mg and T are constant, so is a, and you can use
vf=(2ax)1/2
What is a? a=g-T/m according to Newton's third law. So
vf=(2gx-2Tx/m)1/2, which agrees with what I showed already.
This advice isn't entirely correct. I didn't take into account the pulley's mass.

Try solving these for a.
m1a=m1g-T1
m2a=T2
(T2-T1)R=Ia/R=(1/2)MRa

DocAl's approach might work also, but I would avoid that direction simply because I don't feel comfortable using it.

[ShawnD]

Originally posted by formulajoe
i got 16.3 m/s as a final velocity.
is that even close?

Not even close. I can't give you the real answer since I honestly don't know (I don't know how circular acceleration works) but I can ballpark it.

here's the grade 11 style ballpark:

force down from gravity:
F = ma
F = (2)(9.81)
F = 19.62

acceleration of the 2 blocks (not including circular acceleration on the pully):
a = F/m
a = 19.62 / (2 + 4)
a = 3.27

final velocity:
Vf^2 = Vi^2 + 2ad
Vf^2 = 0 + 2(3.27)(2.5)
Vf^2 = 16.35
Vf = 4.0435 m/s

Remember, THIS IS JUST A BALLPARK FIGURE, you have to factor in the circular acceleration. Your final answer should be LESS than 4.0435

[jamesrc]

Originally posted by ShawnD
I can't give you the real answer since I honestly don't know (I don't know how circular acceleration works) but I can ballpark it.


If you're interested in how to handle the inertia of the pulley, try to follow StephenPrivitera's latest post, though that may be difficult if you haven't learned about torque yet. Otherwise, what you said is true.

There is a lot to be said, however, for the simplicity of DocAl's suggested approach. If you consider the conservation of energy of the entire system (2 masses + pulley), the problem reduces to one line (or two if you write out the geometric constraint as a separate step).
Ei = Ef
m1gΔh = .5*(m1v2 + m2v2 + Iω2)
with m1 = 2kg, m2 = 4kg, Δh = 2.5 m, and I = pulley inertia about its axis = .5MR2, and ω = v/R (relating the pulley rotational speed to the translational speed of the blocks). From that, you immediately solve for v; internal forces are not of interest.

[StephenPrivitera]

Originally posted by jamesrc
There is a lot to be said, however, for the simplicity of DocAl's suggested approach. If you consider the conservation of energy of the entire system (2 masses + pulley), the problem reduces to one line (or two if you write out the geometric constraint as a separate step).
Ei = Ef
m1gΔh = .5*(m1v2 + m2v2 + Iω2)
with m1 = 2kg, m2 = 4kg, Δh = 2.5 m, and I = pulley inertia about its axis = .5MR2, and ω = v/R (relating the pulley rotational speed to the translational speed of the blocks). From that, you immediately solve for v; internal forces are not of interest.
This is genius. I'm putting it on my wall.