(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform thin rod of mass M and length L is supported horizontally by two supports, one at each end. The acceleration of gravity, g, is constant and in the downward direction. At time t=0 the left support is removed.

Find the downward acceleration of the center of mass at t=0 in terms of M, L and g. Also, find the angular acceleration and the force exerted by the remaining support.

2. Relevant equations

[tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]

[tex] |\tau| = I \alpha [/tex]

3. The attempt at a solution

To calculate the downward acceleration, I first calculate the torque.

[tex] \vec{\tau} = \frac{-L}{2}\hat{y} \times (-M g) \hat{z} = \frac{LMg}{2}\hat{x} [/tex]

Equate this torque with the moment of inertia times the angular acceleration, and solve for angular acceleration.

[tex] \alpha = \frac{\tau}{I} = \frac{\frac{L}{2} M g}{\frac{1}{3} M L^2}=\frac{3}{2}\frac{g}{L} [/tex]

Then, solve for the acceleration,

[tex] a = r \alpha = \frac{L}{2} \cdot \frac{3}{2}\frac{g}{L} = \frac{3}{4} g [/tex]

Find the force by F=ma,

[tex] F = m a = M \frac{3}{4} g [/tex]

Where have I gone wrong here? Thanks!

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