kdm06 said:
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,
w1/O =
w1/G = 20
k rad/s. Here, 1 refers to gear 1 and G refers to ground reference. But, A is not fixed. Thus,
w2/A = -10
k rad/s, which is not equal to
w2/G. Now, we are required to find
wA/O, angular velocity of A w.r.t. O (which will also be equal to
wA/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.
vP1/O =
vP2/O.
Now,
vP1/O =
w1/O X
rP1/O.
And, as
vP2 =
vA/O +
vP2/A,
=>
vP2/O =
wA/O X
rA/O +
w2/A X
rP2/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,
vP1/O = 6
j m/s,
and,
vP2/O = (O.5*w
A/O + 2)
j m/s.
Wherefrom, w
A/O = 8 rad/s.
Thus,
wA/O = 8
k rad/s.
And,
vA/O =
wA/O X
rA/O = 4
j m/s.
Thus, v
A/O = 4 m/s.
Note: Try to put up clear questions.