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Re: Atwood's machine problem



Thank you all! Being the lone Physics person on campus, I appreciate
all help (and patience) from those on the list.

Peter Schoch
(whose 15 month old son is cutting molars, and keeping his parents sleep
deprived.)

Herb Schulz wrote:

On 10/1/04 2:27 PM, "Peter Schoch" <pschoch@NAC.NET> wrote:



Hello all,

I just got a question I can't answer, and could use some help.

A student got this problem out of a book (didn't tell which one) and asked
me, and I 'clutched':

An Atwood's machine is in perfect balance, with equal weights on both
sides. One of the weights is a monkey, and the other is a stack of
bananas. The monkey begins to climb at a constant speed. What happens to
the bananas; ie, do they rise, fall, or stay stationary.

My first inclination was to use CM and say that if the monkey climbs the
bananas must fall to keep the CM stationary. However, the book he Xeroxed
this from, said the bananas rise with the monkey.

Maybe it's just because it's Friday, but I don't see the 'mechanism' for
that explanation. Can somebody help?

Thanks,
Peter Schoch
(sleep deprived with a 15 month at home)




Howdy,

Consider the system to be the Monkey, Bananas, Rope (assumed massless) and
Pulley (assumed massless and frictionless and rigidly attached to a wall).
When the monky pulls down on the rope there ends up being anet externaL
force on the system through the pivot of the Pulley (i.e., P-2mg not = 0) so
the CM of the system is free to move. However, the net external torque on
the system measured about the pivot is zero so the Angular Momentum of the
System ABOUT THE PIVOT must be conserved and it was zero before. Therefore
the Bananas must have the same velocity (m and r are the same) at any given
instant as the monkey so the two Angular Momentum Vectors add to give zero.

Good Luck,

Herb Schulz
(herbs@wideopenwest.com)