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Fw: Mechanical Vibration Teaser Problem

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I got an answer similar to Steve Smith's, but a bit different:

w=sqrt(uk*g/32/L), so for L=2.0 cm, uk=.4 I get w=2.47 to go from initial
disp x, to -x and back to x.

How about letting us know the *real* answer so I can let this one go.

T. Eric Gillham PE
GK2 Inc.
PO Box 3207  Agana, Guam  96932
Email - gk2(--nospam--at)
Ph:  (671) 477-9224
Fax: (671) 477-3456

-----Original Message-----
From: Eddie Gonzalez <Eagonzal(--nospam--at)ENG.CI.LA.CA.US>
To: seaint(--nospam--at) <seaint(--nospam--at)>
Date: Friday, January 22, 1999 11:18 AM
Subject: Mechanical Vibration Teaser Problem

>One of our interns had the following mechanical vibration problem to solve.
>thought maybe some of you on the server would get a kick in trying to solve
>-- take you back to your college years.  Its a text book problem (Single
>of Freedom System) so keep it simple.
>A uniform bar with mass "m" lies symmetrically across two rapidly rotating,
>fixed rollers, A and B, with distance L=2.0 cm between the bar's center of
>and each roller.  The rollers, whose direction of rotation are shown in the
>figure, slip against the bar with coefficient of kinetic friction uk =
>Suppose the bar is diplaced horizontally by a distance X, and then
>Q: What is the angular frequency    w     of the resulting horizontal
>harmonic (back and forth) motion of the bar?
>                                     |<------L------>|<-------L------->|
>                           ==============M==============
>                                     O
>                             (clockwise                 (counter-clockwise
>                                 rotation)
>assume: Prior to displacement, bar is stable, with the force of one
>roller couteracting the force of the other.
>Hints:  D' Alambert's Principle, Sum of F = ma
>ed gonzalez