Register to reply 
Kinetics of rigid body 
Share this thread: 
#1
Nov1109, 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/sunterclockwise. 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 r_{B/A}omega^{2}*r_{B/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... 


#2
Nov1109, 07:31 AM

Sci Advisor
HW Helper
Thanks
P: 26,157

Hi yoamocuy!
(have a theta: θ and an alpha: α and a sigma: ∑ ) Yes, use x and y coordinates, but do everything as a function of θ (in particular, this will give you a relation between a_{A }and a_{B}). (no need to do anything fancy like rotating coordinates ) 


#3
Nov1109, 04:24 PM

P: 40

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(θ_{1})+Mgtan(θ_{2})+Ax=3.6a θ_{1}=15^{o} θ_{2}=30^{o} 


#4
Nov1109, 04:36 PM

Sci Advisor
HW Helper
Thanks
P: 26,157

Kinetics of rigid body
Start by finding the equation of motion … use conservation of energy. 


#5
Nov1109, 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*ω^{2}+mgh where I=(1/3)*3.6*1.2^{2}, and h=0.4392m, but I don't think I have another half of the equation.



#6
Nov1109, 05:15 PM

P: 40

O, could I use the equation d=vf^{2}vo^{2}/2*a? So id have 30^{o}=ωf^{2}ωo^{2}/2*alhpa
ωf=2, ωo=0 and 30^{o}=pi/6=0.523599 So plugging in all values and solving for alpha would give me alpha=3.82 rad/s^{2} 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? 


#7
Nov1109, 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? 


#8
Nov1209, 08:53 AM

Sci Advisor
HW Helper
Thanks
P: 26,157

Good morning!
(except that the centre of mass is moving, so KE = KE_{trans} + KE_{rot}, so you have to add on 1/2 mv_{c.o.m}^{2}) 


#9
Nov1209, 05:29 PM

P: 40

Alright I got the answer, thanks for your help.



Register to reply 
Related Discussions  
Kinetics of a rigid body  Introductory Physics Homework  0  
Rigid Body Kinetics  Introductory Physics Homework  2  
Rigid body kinetics  Introductory Physics Homework  1  
Planar Kinetics of a Rigid Body .. Please Help :S  Introductory Physics Homework  1  
Planar Kinetics of a Rigid Body  Introductory Physics Homework  6 