- #1
Karol
- 1,380
- 22
Homework Statement
An upright standing rod of mass m and length 2l receives a small push at the top and falls.
The floor is smooth. the height of the center of mass at any given time is h, according to the drawing.
Show that the velocity of it's center of mass is:
[tex]V=(l-h)\sqrt{6g(l+h)/(4l^2-3h^2)}[/tex]
Homework Equations
[tex]M=I\alpha[/tex]
The Attempt at a Solution
Because all forces are vertical, I assume the center of mass will move directly vertical, under it's initial location, as in the drawing.
The horizontal distance between the reaction of the floor F and the center of mass is:
[tex]x=lcos\theta=l \sqrt{1-sin^2\theta}=l \sqrt{1-\frac{h^2}{l^2}}=\sqrt{l^2-h^2}[/tex]
[tex]\left\{\begin{array}{l}mg-F=ma\\F\sqrt{l^2-h^2}=\frac{4}{3}ml^2\alpha
\\a=\alpha l\end{array}[/tex]
From those:
[tex]a=\frac{3g\sqrt{l^2-h^2}}{4l+3\sqrt{l^2-h^2}}[/tex]
And now:
[tex]V=\int_{l}^{h}\frac{3g\sqrt{l^2-h^2}}{4l+3\sqrt{l^2-h^2}}dh[/tex]
Which, I tried, doesn't yield the required result.