# Homework Help: Kinetics of rigid body

1. Nov 11, 2009

### yoamocuy

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
$$\Sigma$$F=ma
$$\Sigma$$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...

#### Attached Files:

• ###### Untitled.jpg
File size:
5 KB
Views:
83
2. Nov 11, 2009

### tiny-tim

Hi yoamocuy!

(have a theta: θ and an alpha: α and a sigma: ∑ )
(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 )

3. Nov 11, 2009

### yoamocuy

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

#### Attached Files:

• ###### Untitled.jpg
File size:
5.6 KB
Views:
103
Last edited: Nov 11, 2009
4. Nov 11, 2009

### tiny-tim

Sorry, I have no idea what you're doing.

Start by finding the equation of motion …

use conservation of energy.​

5. Nov 11, 2009

### yoamocuy

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.

6. Nov 11, 2009

### yoamocuy

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?

7. Nov 11, 2009

### yoamocuy

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.

8. Nov 12, 2009

### tiny-tim

conservation of energy

Good morning!
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)

9. Nov 12, 2009