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Hello,
I need to confirm my solution to this problem: A thin rod of length l stands vertically on a table. The rod begins to fall, but its lower end does not slide. (a) Determine the angular velocity of the rod as a function of the angle \phi it makes with the tabletop. (b) What is the speed of the tip of the rod just before it strikes the table.
For (a), I found the moment of inertia I for the rod as well as the torque. Then, I equated the expression I found for the torque with \tau = I\alpha and solved for \alpha. Knowing \alpha, I then calculated the angular velocity \omega from it with some calculus.
For (b), the answer would be
v = R\omega(0)
using the \omega I found in (a).
I was going to do the problem by simplifying the rod using center of mass concepts, but it seems I can't do this with rotational motion.
Can you use the Lagrangian formalism? I'll work through the problem using it and post what I get.
Huh. The Lagrangian formalism leads to a nonlinear diff. eq..
OK, I adjusted the formalism a little bit, and I got the general expression for part a to be of the form:
ω = (+/-)√{ A - B sinφ }.
Did you get something of this form for part a? I think you probably would using the technique you mentioned.
For part b, you should just be able to plug in φ = 0 (or π) as you have done. I'm assuming that, by R, you mean l.
For (a), I found the moment of inertia I for the rod as well as the torque. Then, I equated the expression I found for the torque with \tau = I\alpha and solved for \alpha. Knowing \alpha, I then calculated the angular velocity \omega from it with some calculus.
No need for calculus or advanced methods. For (a) use conservation of energy. The rod can be viewed as being in pure rotation, so its KE is just the rotational KE about the pivot point. This will give an answer for ω as a function of φ.
As you noted, for (b) just plug in φ = 0 to find ω. The speed of the tip is v = ωl.
No need for calculus or advanced methods. For (a) use conservation of energy. The rod can be viewed as being in pure rotation, so its KE is just the rotational KE about the pivot point. This will give an answer for ω as a function of φ.
This problem came from a section before rotational kinetic energy is introduced so... . Anyways, let me check my solution using this method.
Let m be the mass of the rod. Using conservation of energy yields
mgl/2 = mgl\sin \phi /2 + I{\omega}^2/2
where I = ml^2/3. Solving for \omega gives
\omega = \sqrt{\frac{3g(1 - \sin \phi)}{l}}
Thank goodness for energy methods.
I'm assuming that, by R, you mean l.
Right. My mistake.
e(ho0n3,
Yeah, that's how I did it. Sorry about my rambling. There is one subtle issue. If you tip the rod over, then you will probably give it some initial kinetic energy as well. The problem doesn't specify, so I would just assume a negligible initial kinetic energy.
e(ho0n3,
Yeah, that's how I did it. Sorry about my rambling. There is one subtle issue. If you tip the rod over, then you will probably give it some initial kinetic energy as well. The problem doesn't specify, so I would just assume a negligible initial kinetic energy.
Right. It seemed funny to me that a rod would just decide to fall on its own, but for the purposes of this problem I guess I'll have to assume this.
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