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Kinetics of rigid body

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yoamocuy
#1
Nov11-09, 07:13 AM
P: 40
1. The problem statement, all variables and given/known data
The small end rollers of the 3.6 kg uniform slender bar are constrained to move in the slow, which lies in a vertical plane. A the instant theta=30 the angular velocity of the bar is 2 d/sunter-clockwise. Determne the angular acceleration of the bar,k the reaction at A and B, and the acceleration of point A and B under the action of the 26 N force P. Neglect the friction and the mass of the small rollers.


2. Relevant equations
[tex]\Sigma[/tex]F=ma
[tex]\Sigma[/tex]Mp=I*alpha

3. The attempt at a solution
Ok so I'm having problems starting this problem. I was thinking that I could set my coordinate so that the x axis is in line with the 26N force P and do a sum of the forces in the x and y direction and a sum of the moments, but I'm not sure how that would work since I have 2 seperate accelerations at point A and B. Would I need to use the equation aB=aA+alpha X rB/A-omega2*rB/A somehow? I think once I figure out how to get started on this question I'll be able to figure out how to solve the rest of it...
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tiny-tim
#2
Nov11-09, 07:31 AM
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Hi yoamocuy!

(have a theta: θ and an alpha: α and a sigma: ∑ )
Quote Quote by yoamocuy View Post
The small end rollers of the 3.6 kg uniform slender bar are constrained to move in the slow, which lies in a vertical plane. A the instant theta=30 the angular velocity of the bar is 2 d/sunter-clockwise. Determne the angular acceleration of the bar,k the reaction at A and B, and the acceleration of point A and B under the action of the 26 N force P. Neglect the friction and the mass of the small rollers.
(did you mean deg/s ?)

Yes, use x and y coordinates, but do everything as a function of θ (in particular, this will give you a relation between aA and aB).

(no need to do anything fancy like rotating coordinates )
yoamocuy
#3
Nov11-09, 04:24 PM
P: 40
Quote Quote by tiny-tim View Post
(did you mean deg/s ?)
I meant rad/s :P

I attached another diagram with an axis drawn in. So, when you say to leave it in terms of θ, do you
mean something like this?

∑Fx=26cos(θ1)+Mgtan(θ2)+Ax=3.6a

θ1=15o
θ2=30o
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tiny-tim
#4
Nov11-09, 04:36 PM
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Kinetics of rigid body

Quote Quote by yoamocuy View Post
do you
mean something like this?

Sum of Fx=26cos(θ1)+Mgtan(θ2)+Ax=3.6a
Sorry, I have no idea what you're doing.

Start by finding the equation of motion
use conservation of energy.
yoamocuy
#5
Nov11-09, 04:53 PM
P: 40
How would use the conservation of energy for this problem? Isn't the rod at the same position for the entire problem? I can set up 1 side of the conservatin of energy to be equal to .5*I*ω2+mgh where I=(1/3)*3.6*1.22, and h=0.4392m, but I don't think I have another half of the equation.
yoamocuy
#6
Nov11-09, 05:15 PM
P: 40
O, could I use the equation d=vf2-vo2/2*a? So id have 30o=ωf2-ωo2/2*alhpa
ωf=2, ωo=0 and 30o=pi/6=0.523599

So plugging in all values and solving for alpha would give me alpha=3.82 rad/s2

Then I thnk I could do a sum of the moments and forces to find the reaction forces at a and b but how would I find the accelerations?
yoamocuy
#7
Nov11-09, 06:22 PM
P: 40
Ok, I used that value for alpha and tried to do a sum of the moments, sum of the forces in the x direction, and some of the forces in the y direction to solve for the forces at each roller. After much geometry I ended up with 3 equations that look like this:

1) ∑M=.424B-.52A+11.04=6.6

6.6 is the product I*alpha, A is the force at the top roller and B is the force at the bottom roller.

2)∑Fx=26*cos(15)+B*sin(15)+A=3.6*a

a=acceleration of entire rigid body?

3)∑Fy=-26*sin15+B*cos(15)-3.6*9.81=3.6*a

note: I used the coordinate system drawn on my 2nd diagram.

So after solving for all 3 variables I got numbers that don't seem logical... so basically I'm kinda stuck at this point. I'm not sure if I'm even approaching this problem in the right manner.

Any advice?
tiny-tim
#8
Nov12-09, 08:53 AM
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Good morning!
Quote Quote by yoamocuy View Post
How would use the conservation of energy for this problem? Isn't the rod at the same position for the entire problem? I can set up 1 side of the conservatin of energy to be equal to .5*I*ω2+mgh where I=(1/3)*3.6*1.22, and h=0.4392m, but I don't think I have another half of the equation.
There is no other side it's just 1/2 Iω2 + mgh = constant.

(except that the centre of mass is moving, so KE = KEtrans + KErot, so you have to add on 1/2 mvc.o.m2)
yoamocuy
#9
Nov12-09, 05:29 PM
P: 40
Alright I got the answer, thanks for your help.


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