Why Is This Dynamics Problem So Challenging?

[Saladsamurai]

!Desperate!This Dynamics Problem is Killing ME!

Homework Statement

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.

Can someone please help me out here?

Thanks

 

[Saladsamurai]

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?

 

[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.

 

[HallsofIvy]

Homework Statement

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.

Can someone please help me out here?

Thanks

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.

 

[Saladsamurai]

....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.

 

[Saladsamurai]

...

 

[Saladsamurai]

Does anyone even look in this forum?

 

[SEG9585]

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.

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
800 mm / (60mm radius)/pi * (2*pi radians/turn) = 26.67 rad/sec

Now that you have omegaC, use that rotation rate to find the distance gear B travels.