Question

The bar shown above is pivoted about one end and is initially at rest in a vertical position. The bar is displaced slightly and as it falls it makes an angle theta with the vertical at any given time, as shown above.

Answer 1

The answers are D and E respectively.

Answer 2

@Michele_Laino @IrishBoy123 I understand that the angular acceleration is \[\frac{ a }{ r }\] and the linear acceleration in this case would be the acceleration due to gravity, so a = g.

Answer 3

Which is constant. That is why I think that in this case the angular acceleration would be constant and the answer would be A but I am wrong. I would like to know why?

Answer 4

we have to apply the second cardinal equation of dynamics

Answer 5

the external force acting on our bar is the weight force of the same bar

Answer 6

we neglect all friction forces

Answer 7

Next we have to compute the moment of the weight force with respect the pivoting point

Answer 8

please wait, I compute that moment of weight force

Answer 9

here is my computations:
\[\Large {{\mathbf{M}}^W} = {\mathbf{OG}} \times m{\mathbf{g}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{x}}&{\mathbf{y}}&{\mathbf{z}} \\
{\frac{L}{2}\sin \theta }&{\frac{L}{2}\cos \theta }&0 \\
0&{ - mg}&0
\end{array}} \right| = - mg\frac{L}{2}{\mathbf{z}}\]

Answer 10

so applying the second cardinal of dynamics, I can write:
\[\Large I\alpha = - mg\frac{L}{2}\sin \theta \]
where I is the moment of inertia of the bar with respect to the pivoting point, and L is its length.
Furthermore I have made a typo, since the right formula for moment of weight is:
\[\Large{{\mathbf{M}}^W} = {\mathbf{OG}} \times m{\mathbf{g}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{x}}&{\mathbf{y}}&{\mathbf{z}} \\
{\frac{L}{2}\sin \theta }&{\frac{L}{2}\cos \theta }&0 \\
0&{ - mg}&0
\end{array}} \right| = - mg\frac{L}{2}\sin \theta {\mathbf{z}}\]
here \alpha is the angular acceleration

Answer 11

oops...
\[\Large \begin{gathered}
{{\mathbf{M}}^W} = {\mathbf{OG}} \times m{\mathbf{g}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{x}}&{\mathbf{y}}&{\mathbf{z}} \\
{\frac{L}{2}\sin \theta }&{\frac{L}{2}\cos \theta }&0 \\
0&{ - mg}&0
\end{array}} \right| = \hfill \\
\hfill \\
= - mg\frac{L}{2}\sin \theta {\mathbf{z}} \hfill \\
\end{gathered} \]

Answer 12

furthermor G is the center of gravity of our bar:

Answer 13

The force mg is applied at G right?

Answer 14

In all cases a force is always applied at the center of mass?

Answer 15

that's right!