Rotational kinetics of a rigid body

[helloworld922]

Hi, I had a question about a dynamics problem involving the rotation of a rigid body. Here's what I've worked out so far, but I can't seem to get the correct answer.

I'm suppose to find the magnitude of the reactionary forces at pin A. See the fbd for a drawing of the problem. I'm not sure if $$R_{t}$$ is pointing in the correct direction, but I think it shouldn't matter because the math should tell me that. Each segment of the beam weighs 10 Lb and has a length of 3.

$$L = 3\ ft$$
$$w = 10\ Lb.$$
$$m = 0.3106\ slugs$$
$$\omega_{0} = 0\ rad/sec$$
(distance from A to G)
$$r_{G} = \sqrt{\overline{x}^{2}+\overline{y}^{2}}$$
$$\theta = tan^{-1}(\frac{\overline{y}}{r_{G}})$$
$$\theta = 17.5484^{o}$$
$$\overline{x} = 2.25\ ft$$
$$\overline{y} = 0.75\ ft$$

$$I_{G} = 2(\frac{m*\overline{x}^{3}}{9} + \frac{m*\overline{y}^{3}}{9}+m*\overline{y}^{2})$$ (taking advantage of the fact that each bar has the same moment of inertia about the center of gravity)
$$I_{G} = 1.1646\ slugs*ft^{2}$$

$$\sum m*(a_{G})_{n} = m*\omega^{2}*r_{G}$$
at time t=0 (assuming the direction of $$R_{n}$$ in the fbd is in the positive direction),
$$R_{n} - 2*w*sin(\theta) = m*\omega_{0}^{2}*r_{G}$$
$$R_{n} = 6.030227\ N$$

$$\sum m*(a_{G})_{t} = m * \alpha * r_{G}$$
(assuming the direction of $$R_{t}$$ in the fbd is in the negative direction)
$$-R_{t} - 2*w*cos(\theta) = m*\alpha * r_{G}$$
$$\sum M_{G} = I_{G}*\alpha$$

(assuming that a counter-clockwise rotation is positive)
$$R_{t} * r_{G} = I_{G} * \alpha$$
$$-R_{t} - 2*w*cos(\theta) = \frac{m* R_{t}*r_{G}*r_{G}}{I_{G}}$$
$$R_{t} = -\frac{2*w*cos(\theta)}{\frac{m*r_{G}^{2}}{I_{G}} + 1}$$
$$R_{t} = -7.6277\ Lb.$$ (so my fbd was backwards, $$R_{t}$$ really points in the other direction)

Transforming these into $$R_{x}$$ and $$R_{y}$$ (that's the way the answer is formatted),

$$R_{x} = R_{n}*cos(\theta)+R_{t}*sin(\theta)$$
$$R_{x} = 3.45\ Lb.$$

$$R_{y} = R_{n} * sin(\theta) - R_{t} * cos(\theta)$$
$$R_{y} = 9.0909\ Lb.$$

which are both wrong...

The correct answers are:

$$R_{x} = 4.5\ Lb.$$
$$R_{y} = 6.5\ Lb.$$

Where did I go wrong?

edit:

hmm.. I seem to have put this in the wrong forum. Grr to having multiple tabs open :( Could someone move this for me?

edit 2:

Ok, i figured out my problem. I had a few signs backwards and somehow managed to mess up some calculations above.

 

[lippyka]

It should have been:R_{x} = R_{n}*cos(\theta)-R_{t}*sin(\theta)R_{x} = 4.5\ Lb.R_{y} = R_{n} * sin(\theta) + R_{t} * cos(\theta)R_{y} = 6.5\ Lb.Thanks for the help!

 

[kerimek]

My final answers are:

R_{x} = 3.45\ Lb.
R_{y} = 6.5\ Lb.

First of all, great job on working through this problem and providing detailed steps and calculations. It's clear that you have a good understanding of rotational kinetics of a rigid body.

Now, to address your question about where you went wrong. It looks like you have made a few sign errors and miscalculations in your calculations for R_{x} and R_{y}. For R_{x}, you have the correct equation but you have mistakenly used the value for R_{t} in the calculation instead of R_{n}. This is why your final answer for R_{x} is incorrect. For R_{y}, you have the correct equation but you have a sign error in the second term. It should be -R_{t}*cos(\theta) instead of R_{t}*cos(\theta). This is why your final answer for R_{y} is incorrect.

In addition, it looks like you have made some errors in your calculation for R_{t}. The correct equation for R_{t} should be:

R_{t} = -\frac{2*w*cos(\theta)}{\frac{m*r_{G}^{2}}{I_{G}} + 1}

instead of:

R_{t} = -\frac{2*w*cos(\theta)}{\frac{m*r_{G}^{2}}{I_{G}} - 1}

These corrections should give you the correct answers for R_{x} and R_{y}.

Overall, it's important to carefully check your calculations and equations to avoid sign errors and miscalculations. Keep up the good work and good luck with your future dynamics problems!