Relative Velocity problem: 2 pulleys riveted together

[bkw2694]

Homework Statement

As shown in the picture, two pulleys are riveted together with a known velocity. I need to find the velocity of two other points.

Homework Equations

V_A = VB + <ω> x <BA>

V = ωr

The Attempt at a Solution

I've honestly been trying to figure this one out for over a week, and I'm sure I'll kick myself when I realize how easy it is, but I'm trying to solve by:

V_A = (r_OA)*(ω_OA), using V_A = .9 m/s; r_OA = 90mm, which gives me ω_OA = 10 rad/sec (which I assume ω is equal on the entire pulley since it's riveted together).

Then I used the relative velocity vector formula for A and O:

V_A = V_O + <-10 rad/sec k> x <-.09 m i>
which gives me: .9 m/s j = V_O j + .9 m/s j. Solving with this gives me V_O = 0, which is incorrect. I tried solving for B the same way.

The actual answers are V_O = .6 m/s, and V_B = .849 m/s.

Thanks in advance!

 

[TSny]

V_A = (r_OA)*(ω_OA)

This would give the speed of A relative to O, not relative to the earth.

 

[bkw2694]

This would give the speed of A relative to O, not relative to the earth.

I thought <w>x<OA> was the speed of A relative to O? I thought the equation was sufficient for finding the actual speed of A? How would I solve for thespeed relative to earth?

 

[TSny]

I thought <w>x<OA> was the speed of A relative to O?

Yes, it's the speed of A relative to O. But point O is not at rest relative to the earth.

How would I solve for the speed relative to earth?

Can you find a point of the pulley that is instantaneously at rest relative to the earth? Hint: Think about problems where you have rolling without slipping.

 

[bkw2694]

Yes, it's the speed of A relative to O. But point O is not at rest relative to the earth. Can you find a point of the pulley that is instantaneously at rest relative to the earth? Hint: Think about problems where you have rolling without slipping.

Ahhhh, I see now. So I just plugged V_O = .18w into the equation and solved for w first. Thanks so much!

 

[TSny]

OK. Or, you can get $\omega$ from $V_A = .27 \omega$, since $V_A$ is given.

 

[bkw2694]

OK. Or, you can get $\omega$ from $V_A = .27 \omega$, since $V_A$ is given.

I see that that definitely works mathematically, but I don't understand why it works by looking at the problem. Why is it .27 instead of .18? I don't understand why the distance is .27 when the velocity vector is .18 m away.

Thanks again!

 

[TSny]

Let point D be the point on the outer rim of the pulley where the string on the right meets the pulley.

Use the fact that $\vec{V}_A =\vec{V}_D +\vec{V}_{A/D}$.

 

[J Hann]

It might help if you consider what happens to a simpler situation where a string raises a single pulley
with one end of the string attached to the ceiling.
In the given problem you have:
V0 = .18 * w where w is the angular velocity
VA = V0 + .09 w
and this is easily solved for w, the angular velocity.