- #1
Metaleer
- 124
- 0
Hey, all!
Browsing around the library for some mechanics books, I happened to come across Manuel Prieto Alberca's Curso de Mecánica Racional. Dinámica. In this book, I managed to find an interesting problem that is also solved, but having talked to other people, I have started to suspect that the reference solution is wrong.
I will now copy the problem statement:
Consider a bar [itex]AB[/itex] of length [itex]\ell[/itex] and mass [itex]m[/itex] that falls down due to gravity, starting from rest at [itex]Ox[/itex]. The point [itex]A[/itex] of the bar is attached to [itex]O[/itex] with a massless string of length [itex]h[/itex]. Determine the angular velocity and velocity of the center of mass right after the string has completely stretched out, that is, reached a length of [itex]h[/itex].
The reference solution goes as follows:
Let [itex](\xi, \eta)[/itex] be the coordinates of the center of mass of the bar and [itex]\theta[/itex] the angle it forms with the horizontal in an arbitrary position.
We have
We wish to determine these values right after the percussion.
We have the constraint
Applying Lagrange's equations we have
And since the constraint is persistent, we have
[tex]\left(\xi - \frac{\ell}{2}\cos \theta \right) \left(\dot{\xi}_1 + \frac{\ell}{2} \dot{\theta}_1 \sin \theta \right) + \left(\eta - \frac{\ell}{2}\sin \theta \right) \left(\dot{\eta}_1 - \frac{\ell}{2} \dot{\theta}_1 \cos \theta \right) = 0[/tex]
Since at the moment of the percussion we have [itex]\theta = 0[/itex], [itex]\xi = \ell/2[/itex], [itex]\eta = h[/itex], we are left with
System that solved yields
[tex]\dot{\xi}_1 = 0 ~ ~ ~ \dot{\eta}_1 = \frac{3}{4}\sqrt{2gh} ~ ~ ~ \dot{\theta}_1 = \frac{3 \sqrt{2gh}}{2\ell}[/tex]
So, having shown the reference solution, the issue raised by people I have to talked to is the fact that the constraint applied is for a suddenly appearing hard bar.
The string isn't stiff, even though it doesn't change length during the moment of impact; it allows a velocity component at [itex]A[/itex] that changes the length of the string.
Thus, an alternative approach suggested by the same person was to include the ceiling the string is attached to at the top (point [itex]O[/itex]), use conservation of energy and momentum and take the limit for infinite mass for the ceiling.
So, what do you guys think? Is the reference solution wrong? Would the alternative approach seem better?
I'm interested in knowing what you think.
Edit:
For moderators: I'm not 100% sure if this goes here or in the HW/Coursework section. It is a problem from a textbook, but I was looking for a section where instead of plain academic assistance, I was expecting an engaging conversation with knowledgeable people on why this method doesn't work, why the other may, etc... If it's in the incorrect section, my apologies to the moderation staff and administrators. :)
Browsing around the library for some mechanics books, I happened to come across Manuel Prieto Alberca's Curso de Mecánica Racional. Dinámica. In this book, I managed to find an interesting problem that is also solved, but having talked to other people, I have started to suspect that the reference solution is wrong.
I will now copy the problem statement:
Consider a bar [itex]AB[/itex] of length [itex]\ell[/itex] and mass [itex]m[/itex] that falls down due to gravity, starting from rest at [itex]Ox[/itex]. The point [itex]A[/itex] of the bar is attached to [itex]O[/itex] with a massless string of length [itex]h[/itex]. Determine the angular velocity and velocity of the center of mass right after the string has completely stretched out, that is, reached a length of [itex]h[/itex].
The reference solution goes as follows:
Let [itex](\xi, \eta)[/itex] be the coordinates of the center of mass of the bar and [itex]\theta[/itex] the angle it forms with the horizontal in an arbitrary position.
We have
[tex]T = \frac{1}{2}m(\dot{\xi}^2 + \dot{\eta}^2)+ \frac{1}{2}\cdot\frac{1}{12}m\ell^2 \dot{\theta}^2[/tex]
Right before the percussion, we have the following values
[tex]\dot{\xi_0} = 0, ~ \dot{\eta_0} = \sqrt{2gh}, ~ \dot{\theta_0} = 0[/tex]
We wish to determine these values right after the percussion.
We have the constraint
[tex]\left(\xi - \frac{\ell}{2} \cos \theta\right)^2 + \left(\eta - \frac{\ell}{2} \sin \theta\right)^2 = h^2[/tex]
Applying Lagrange's equations we have
[tex]m \dot{\xi}_1 - 2\lambda \left(\xi - \frac{\ell}{2}\cos \theta \right) = 0[/tex]
[tex]m \dot{\eta}_1 - m \sqrt{2 g h} - 2\lambda \left(\eta - \frac{\ell}{2}\sin \theta \right) = 0[/tex]
[tex]\frac{m\ell^2}{12} \dot{\theta}_1 - \lambda\left[ \left(\xi - \frac{\ell}{2}\cos \theta \right) \ell \sin \theta - \left(\eta - \frac{\ell}{2}\sin \theta \right) \ell \cos \theta \right] = 0[/tex]
[tex]m \dot{\eta}_1 - m \sqrt{2 g h} - 2\lambda \left(\eta - \frac{\ell}{2}\sin \theta \right) = 0[/tex]
[tex]\frac{m\ell^2}{12} \dot{\theta}_1 - \lambda\left[ \left(\xi - \frac{\ell}{2}\cos \theta \right) \ell \sin \theta - \left(\eta - \frac{\ell}{2}\sin \theta \right) \ell \cos \theta \right] = 0[/tex]
And since the constraint is persistent, we have
[tex]\left(\xi - \frac{\ell}{2}\cos \theta \right) \left(\dot{\xi}_1 + \frac{\ell}{2} \dot{\theta}_1 \sin \theta \right) + \left(\eta - \frac{\ell}{2}\sin \theta \right) \left(\dot{\eta}_1 - \frac{\ell}{2} \dot{\theta}_1 \cos \theta \right) = 0[/tex]
Since at the moment of the percussion we have [itex]\theta = 0[/itex], [itex]\xi = \ell/2[/itex], [itex]\eta = h[/itex], we are left with
[tex]m \dot{\xi}_1 = 0[/tex]
[tex]m \dot{\eta}_1 - m\sqrt{2gh} - 2 \lambda h = 0[/tex]
[tex]\frac{m\ell^2}{12} \dot{\theta}_1 + \lambda \ell h = 0[/tex]
[tex]\dot{\eta}_1 - \frac{\ell}{2}\dot{\theta}_1 = 0[/tex]
[tex]m \dot{\eta}_1 - m\sqrt{2gh} - 2 \lambda h = 0[/tex]
[tex]\frac{m\ell^2}{12} \dot{\theta}_1 + \lambda \ell h = 0[/tex]
[tex]\dot{\eta}_1 - \frac{\ell}{2}\dot{\theta}_1 = 0[/tex]
System that solved yields
[tex]\dot{\xi}_1 = 0 ~ ~ ~ \dot{\eta}_1 = \frac{3}{4}\sqrt{2gh} ~ ~ ~ \dot{\theta}_1 = \frac{3 \sqrt{2gh}}{2\ell}[/tex]
So, having shown the reference solution, the issue raised by people I have to talked to is the fact that the constraint applied is for a suddenly appearing hard bar.
The string isn't stiff, even though it doesn't change length during the moment of impact; it allows a velocity component at [itex]A[/itex] that changes the length of the string.
Thus, an alternative approach suggested by the same person was to include the ceiling the string is attached to at the top (point [itex]O[/itex]), use conservation of energy and momentum and take the limit for infinite mass for the ceiling.
So, what do you guys think? Is the reference solution wrong? Would the alternative approach seem better?
I'm interested in knowing what you think.
Edit:
For moderators: I'm not 100% sure if this goes here or in the HW/Coursework section. It is a problem from a textbook, but I was looking for a section where instead of plain academic assistance, I was expecting an engaging conversation with knowledgeable people on why this method doesn't work, why the other may, etc... If it's in the incorrect section, my apologies to the moderation staff and administrators. :)
Last edited: