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A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released?

I just used

$$sin(30) = \frac{9.8}{a_{centripetal}}$$

Then I related

$$a_{rad} = a_{centripetal}$$

Is this right? I am looking at University Physics:

$$a_{rad} = v^2 / r$$

Where 9.8 is gravity. But I got none of the answers in the multiple choice ... so I must be doing wrong. Also I haven't used the radius. Any suggestions would be helpful...

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3 Answers

up vote 2 down vote accepted

Always start with a nice clear diagram/sketch of the problem. It all follows from there. Here is a Free Body Diagram I made for you.

FBD

Then you have (the long detailed way):

  1. Sum of the forces on body equals mass times acceleration at the center of gravity. $\sum_i \vec{F}_i = m \vec{a}_C $

    $$ A_x = m a_x \\ A_y - m g = m a_y $$

  2. Sum of torques about center of gravity equals moment of inertia times angular acceleration. $\sum_i \left(\vec{M}_i + (\vec{r}_i-\vec{r}_C)\times\vec{F}_i\right) = I_C \vec{\alpha} $

    $$ A_x \frac{L}{2} \sin(\theta) - A_y \frac{L}{2} \cos(\theta) = I_C \ddot \theta $$

  3. Acceleration of point A must be zero. $\vec{a}_A = \vec{a}_C + \vec{\alpha}\times(\vec{r}_A-\vec{r}_C) + \vec{\omega}\times(\vec{v}_A-\vec{v}_C) $

    $$ a_x + \frac{L}{2} \sin(\theta) \ddot\theta + \frac{L}{2} {\dot\theta}^2 \cos(\theta) =0 \\ a_y - \frac{L}{2} \cos(\theta) \ddot\theta + \frac{L}{2} {\dot\theta}^2 \sin(\theta) =0 $$

    Now you can solve for $a_x$, $a_y$ from 3. and use those in 1. to get $A_x$,$A_y$. Finally use 2. to solve for $\ddot\theta$

Or do the shortcut of finding the applied torque on A and applying it to the effective moment of inertia about the pivot $I_A = I_C + m \left(\frac{L}{2}\right)^2 $ to get

$$ \ddot\theta = \frac{m g \frac{L}{2} \cos\theta }{ I_C + m \left(\frac{L}{2}\right)^2 } $$

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You have been given a rod, and a force along the rod, and you want angular acceleration, these three ingredients point to two equations, but in your mind you should jump to one above the other $$ \tau = I\alpha $$ This says that the sum of the torques equal the moment of inertia time the angular acceleration. For the moment of inertia, in your book there are some lists of moments of inertia for various circumstances, find the one that corresponds to a rod being rotated about one end, or look it up on wikipedia. The torque is going to come from this equation $$ \tau=Fr\sin\theta $$ Now the only thing ``pulling'' on this rod is the force of gravity, so $F=mg$. Now ask yourself where it is pulling on it, and it turns out at $r=L/2$, half way up the rod at the center of mass. Now for $\theta$, since the force of gravity is straight down, the rod, the force of gravity, and the ground make a right triangle. We are given one of the angles, $30^{\circ}$, so we know the other since one of them is $90$. The the angle $\theta$ is defined as the angle between the force and the vector $r$ (in our case the angle between the force and the rod), which must be $\theta=60^{\circ}$. Then $$ Fr\sin\theta=mg\frac{L}{2}\sin60=I\alpha\implies \alpha=\frac{mgL\sin60}{2I} $$

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Where did the $L$ go in the final equation? –  ja72 Oct 28 '12 at 20:44
    
lol, i just forgot it. I put it back in. –  kηives Oct 28 '12 at 20:45
    
Also, your $I$ how is it defined? Is it moment of inertia about the CG, or about the pivot? –  ja72 Oct 28 '12 at 20:46
    
As you can see, I didn't say how to compute $I$, just to look it up in either the text or online as the $I$ about the pivot. –  kηives Oct 28 '12 at 21:56
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As you've been given an uniform rod of some length and an angle of drop, it behaves in a similar way to that of a compound pendulum. Have a look at Wiki for an enormous drop angles relative to horizontal axis. In such a case, you've Moment of Inertia acting via gravity $mg$. You know that, the angular acceleration $\alpha$ is related to the moment of inertia $I$ as $$\tau=I\alpha$$

Also, the moment of inertia for something like your rod could be derived using parallel axes theorem and is given by $I=ml^2/3$. As it's a frictionless pendulum which keeps on swinging forever (unless there's air friction), this problem wouldn't be a problem...

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Since the moment of inertia $I$ is proportional to the mass $m$, mass will cancel out of the equation, and its specification is not required. –  Art Brown Oct 28 '12 at 18:17
    
@ArtBrown: I believe, the cancelling out arrives only if we expand $I$ using Parallel axes theorem. Isn't it? –  Waffle's Crazy Peanut Oct 28 '12 at 19:12
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In particular, the moment of inertia of a thin rod (length $l$, mass $m$) about an axis through its end, perpendicular to the rod's length, is $ml^2/3$. –  Art Brown Oct 28 '12 at 23:13
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