- #1
Saketh
- 261
- 2
Problem: An upright rod of mass [tex]m[/tex] and length [tex]L[/tex] is given an impulse [tex]J[/tex] at its base, 45 degrees above the horizontal, which sends the rod flying. What value should [tex]J[/tex] have for the rod to land vertically again (i.e. upright on the same end at which [tex]J[/tex] was applied)?
I'm trying to learn from Feynman's Tips on Physics, so I tried to follow his methods.
First, I thought about it without calculating. If the impulse were applied vertically, then it doesn't matter what its value is - the rod will always land on the same end. If the impulse were applied horizontally, then no impulse would be enough to make the rod land on the same end. Also, the center of mass must move in a parabola throughout the flight.
Armed with this thought process, I moved on to calculations.
[tex]J = mv[/tex]
I know what [tex]m[/tex] is, so I set that aside.
[tex]v_{y}=\frac{J}{m\sqrt{2}}
t_{in flight}=\frac{2v}{g}[/tex]
I found the time that the rod spends in the air.
[tex]\Delta\theta = 4\pi = \omega t[/tex]
This is how much the rod has to spin.
[tex]v = \omega\frac{L}{2}[/tex]
I tried using the rotation equation relating linear and angular motion, but something felt fishy about doing this.
Then I tried using a physical pendulum equation:
[tex]\omega = \sqrt{\frac{mg\frac{L}{2}}{\frac{ML^{2}}{12}}[/tex]
Then I realized I was going nowhere at this point. I am stumped as to what method I am supposed to use. What is the relationship between the initial horizontal velocity and [tex]4\pi[/tex]? That's what I'm looking for right now.
Last edited: