Timeline for Homework help: Finding the min speed for the normal to be zero
Current License: CC BY-SA 3.0
5 events
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| Oct 19, 2015 at 15:53 | comment | added | DoubleOseven | @ Chris Drost. I don't know who you are or what you do but you sure know how to explain things :) Thanks so much for your help!!! | |
| Oct 19, 2015 at 15:49 | comment | added | CR Drost | @Mathguy007 because it doesn't start out with any rotational oomph: up until it crosses over onto the circular part of the track the pipe has been flat, not rotating. Here's a way to see the value of $\omega$: imagine $v \to \infty.$ The pipe only feels the torque $mgL/2$ for a time $L/v,$ picking up an angular velocity $(g/v) mL^2/(2I),$ which goes to zero like $1/v$. We can therefore ask "for very large v, as it flies off, does it have some steady-state rotation which is not tending to zero?" In this case, no: if you let $v$ get large it fires off like a blowgun dart, with no base $\omega$. | |
| Oct 19, 2015 at 15:41 | comment | added | DoubleOseven | Thank you for your detailed answer. I find your way of describing easy to follow. I understand the principle of throwing a ball up into the air as well as the previous example. I don't understand though how that applies to this problem. Could you please comment and explain why in this situation we have omega equal to zero :) | |
| Oct 19, 2015 at 15:20 | vote | accept | DoubleOseven | ||
| Oct 19, 2015 at 15:06 | history | answered | CR Drost | CC BY-SA 3.0 |