Dynamics - Relative velocities in rigid bodies

[kdm06]

Homework Statement

Point O is fixed in space. Gear 1 has an angular velocity of 20k rad/s, and Gear 2 has an angular velocity of -10k rad/s. r1 = 0.3 m and r2 = 0.2 m. What is the angular velocity of the connecting arm OA, and what is the speed of A?

Gear 1 is larger than Gear 2. Gear 1 is connected to point O and Gear 2 is connected to point A.

Homework Equations

vA=vO + (wOA x rA/O)

The Attempt at a Solution

w2=w1 + w2/1
-10k=20k+w2/1
w2/1=-30k

vA=vO+(wOA x rA/O)
vA=0 + -30k x .5i
vA= -6j m/s

The correct answer is 4j m/s. I think I am using the right equations but I am not fully understanding the concept of relative velocity when different bodies are in effect. Can someone please explain this problem to me? Thank you.

 

[kdm06]

Can someone please help me?

 

[saket]

Can someone please help me?

No reply so far could have been because your problem was not very clear! I did have a hard time understanding it. You nowhere mentioned where is A? A figure should have made things easier.

From the question and your attempt at solution, this is what I have been able to figure out: The two gears are touching each other externally! (Note that, they could have been independent as well.)
Furthermore, I have also assumed that the angular velocities provided for each of the gears are about their own axes, not w.r.t. the fixed ground. Of course, since O is fixed, w_1/O = w_1/G = 20k rad/s. Here, 1 refers to gear 1 and G refers to ground reference. But, A is not fixed. Thus, w_2/A = -10k rad/s, which is not equal to w_2/G. Now, we are required to find w_A/O, angular velocity of A w.r.t. O (which will also be equal to w_A/G).

Now to find w_AO, note that there is no slip at contact point. Let us call contact point P. (The point on gear 1 be called P1 and the point on gear 2 be called P2.)
For no-slip condition at P, velocity of P1 must be equal to velocity of point P2.
v_P1/O = v_P2/O.

Now, v_P1/O = w_1/O X r_P1/O.
And, as v_P2 = v_A/O + v_P2/A,
=> v_P2/O = w_A/O X r_A/O + w_2/A X r_P2/A.

{ X stands for vector cross product, and letters in bold refer to vector quantities. }

Now, without loss of any generality, we can assume that P is a point on x-axis. (Note, all angular velocities are in k direction.
Then, v_P1/O = 6j m/s,
and, v_P2/O = (O.5*w_A/O + 2)j m/s.
Wherefrom, w_A/O = 8 rad/s.
Thus, w_A/O = 8k rad/s.

And, v_A/O = w_A/O X r_A/O = 4j m/s.

Thus, v_A/O = 4 m/s.

Note: Try to put up clear questions.

 

[kdm06]

Thank you so much. I'm sorry I didn't have picture because it was confusing. I just don't know how to put a picture on here. I'll try next time. Thank you.