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Re: Correct Answer



I was playing with Interactive Physics when Bowman's message (see below)
arrived. Simulation is justified in this case because it is not easy to
create a uniform B over a large region for a real experiment. The simulation
did show that in the equilibrated state the string is vertical and the
rod with masses is tilted. I did increase air resistance to speed up the
transient oscillatory process. My rope had L=1.5 meters, the rod had L=1 m,
each of the masses (at the end of the rod) was 1 kg and horizontal forces
on masses were 3 N each. The tilt angle was about 20 degrees or so (with
respect to a vertical line).
Ludwik Kowalski
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From: IN%"phys-l@mailer.uwf.edu" 15-OCT-1997 10:36:04.41
To: IN%"phys-l@atlantis.uwf.edu"
CC: IN%"dbowman@tiger.gtc.georgetown.ky.us"
Subj: RE: Correct Answer

Clarence Bennett wrote:
How can the string stay vertical?
If the bar turns, it should stay centered under the top of the string,
unless it is pulled to the side.
But a uniform field would not pull it.

Actually, the string doesn't *stay* vertical. Rather, it ends up vertical.
Initially the string and bar are vertical. After the uniform field is
turned on there is an unbalanced torque on the bar which rotates it. As the
bar rotates the string becomes somewhat tilted as the CM of the bar tends to
stay under the upper string support. When this happens there is an
unbalanced restoring force (as in any displaced pendulum) provided by the
string. This tends to cause the bar/string system to swing back towards the
vertical. Since the system is presumably quite underdamped the bar/string
swings back and forth until dissipation finally causes it to end up with a
vertical string. Meanwhile, the underdamped rotation mode of the bar magnet
in the magnetic field also overshoots its equilibrium orientation and it
undergoes underdamped rotational oscillations (while it is also swinging on
the string). After everything settles down in its equilibrium state the
string is vertical and the bar is tilted at such an angle that the sum of
torques (acting about any specified point on the bar) due to the vertical
supporting force of the string, the bar's distributed gravitational weight,
and the magnetic torque is zero.
David Bowman
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