Kinematics of a Slender Bar with Constrained End Rollers

[yoamocuy]

Homework Statement

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.

Homework Equations

$$\Sigma$$F=ma
$$\Sigma$$Mp=I*alpha

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 separate accelerations at point A and B. Would I need to use the equation aB=aA+alpha X r_B/A-omega²*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...

 

[tiny-tim]

Hi yoamocuy!

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

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 a_A and a_B).

(no need to do anything fancy like rotating coordinates )

 

[yoamocuy]

(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=15^o
θ₂=30^o

 

[tiny-tim]

… 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]

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]

O, could I use the equation d=vf²-vo²/2*a? So id have 30^o=ωf²-ωo²/2*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/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]

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]

conservation of energy

Good morning!

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 = KE_trans + KE_rot, so you have to add on 1/2 mv_c.o.m²)

 

[yoamocuy]

Alright I got the answer, thanks for your help.