# This Dynamics Problem is hard

1. Apr 17, 2008

!!!Desperate!!This Dynamics Problem is Killing ME!!

1. The problem statement, all variables and given/known data
We have ben dealing with General Plane motion of a rigid body and this problem has been getting the best of me for some time now

I know from the text that the answers are $\omega_a=14 rad/s$ $\omega_b=28.8 rad/s$ and $\omega_c=26.7 rads\s$

Now so far this is all I have been able to accomplish. I know that since member DE id rotating about E, the velocity of gear C's center of gravity is

$v_c=r_{DE}\omega_D=.8 m/s$

I also know that this velocity must be equal to $r_c\omega_c\Rightarrow \omega_C=\frac{.8}{.03}=26.667$ This I presume is true since where C makes contact at F can be considered the instantaneous center of zero velocity (IC from now on).

I cannot seem to get this concept to work for the other gears though. I think that where gear C and B meet, their tangential velocities must be equal. Thus using IC to find that velocity,

$(v_t)_c=(v_t)_b=r_{c/IC}*\omega_c=.06*26.667=1.60 m/s$ But now I am lost. I want to find the angular velocities of gears B and A.

Thanks

2. Apr 18, 2008

Okay, can someone clear this up for me....There is a velocity pointing to the right in between gears C and B, let's call it Vp. Now there is also a velocity pointing leftward between gears B and A, let's call it Vp'.... now here is my question: IS Vp=Vp' ?

I am looking at a solution to this problem that the person who wrote it skipped a lot of steps in. It appears from their solution that Vp is not equal to Vp'. I don't see why not?

3. Apr 18, 2008

### Nesha

Maybe, because there are relative and apsolute velocity? You are right that same points have same velocity Vp=Vp' but you have velocity from DE. I have no time in this moment to solve you this problem. I'll try later.

4. Apr 18, 2008

### HallsofIvy

I'm no expert in mechanics or dynamics but here is how I would do this problem:
The outer ring has radius 50+ 40+ 30= 120 mm. Since the arm is rotating at 5 radians/s and there are $2\pi$ radians in the entire circle, the arm will complete $5/2\pi$ of the entire circle in a second. That circle has circumference $240\pi$ mm so the arm covers $(5/2\pi)(240\pi)$= 600 mm/s. The cog with radius 30 mm must cover that same distance in one second. Since it has circumference $60\pi$ mm, that means it must make $600/60\pi= 10/\pi$ complete turns per second. At $2\pi$ radians per turn, that is 20 radians/s.

The other two cogs, with radii of 40 and 50 mm, must also cover 600 mm/s and so you can calculate their angular speeds the same way.

5. Apr 18, 2008

.................I like the approach Halls, however I know that 20 rad/s is not the solution to that gear, as given in OP. I think it is the relative velocities that are screwing me up.

6. Apr 18, 2008

........

7. Apr 19, 2008

Does anyone even look in this forum?

8. Apr 20, 2008

### SEG9585

Small error here -- you have the wrong distance to the end of the link. It should be 50 + 80 + 30 = 160 mm.
So:
160mm * 2*pi = 320*pi
(5 rad/sec) / 2*pi = 800 mm