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...
Answer
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.
Then you have (the long detailed way):
Sum of the forces on body equals mass times acceleration at the center of gravity. $\sum_i \vec{F}_i = m \vec{a}_C $
$$ A_x = m a_x \\ A_y - m g = m a_y $$
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} $
$$ A_x \frac{L}{2} \sin(\theta) - A_y \frac{L}{2} \cos(\theta) = I_C \ddot \theta $$
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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