"Cary Kittrell" <cary at afone.as.arizona.edu> wrote in message
news:ahkcm8$og6$1 at oasis.ccit.arizona.edu...
> In article "Parse Tree" <parsetree at hotmail.com> writes:
> <
> <"Cary Kittrell" <cary at afone.as.arizona.edu> wrote in message
> <news:ahk14h$ib0$1 at oasis.ccit.arizona.edu...> <> In article "John Knight" <johnknight at usa.com> writes:
> <>
> <> <Sqr means square-root of the equation in the parenthesis ().
> <> <So, the resulting velocity would be the same, as the same time is
spent
> <on
> <> <the fall, and the tension would be zero.
> <>
> <> The "resulting velocity would be the same" if both masses were
> <experiencing
> <> the same acceleration the instant of release, but they were not. The
> <> bottom mass was experiencing -2mg downards due to gravity and +2mg
upwards
> <> due to the tension in the spring. The upper mass is experiencing a
> <> now unopposed -mg downwards due to gravity and a -2mg downwards due
> <> to the same spring tension. You figure it out.
> <
> <But they really are experiencing the same acceleration at the instant of
> <release.
>> That is right as far is acceleration due to gravity is concerned,
> as is implied in my statement above. But each body is experiencing
> additional forces due to the spring, so they will no be subject
> to the same accelerations. If you mentally switch off gravity,
> the two bodies will move towards one another with an acceleration
> proportional to the 2mg tension in the spring. If you now switch
> gravity back on, the whole system will accelerate downards at
> 1 g, but this acts equally on the whole system, so you're back
> to considering things in the frame of referrence of the system
> itself -- as Jet implied.
This is not true. If you switch off gravity, then each sphere will stay at
rest. Firstly, you're assuming tension again, and secondly, the tension you
assume exists only because of gravity.
Regardless, you can simulate this using two balls and a string. Just put
them on a table and attach them with some string. Then pull on them and
release. They don't move together with a force proprotionational to how
much you pulled them apart.
> <Also, you're assuming the value of the unknown.
> <
>> Um, beg pardon? Assuming the value of what unknown? If you mean
> the spring tension, I simply said that the /initial/ spring
> tension is 2mg, because the lower mass is being pulled downwards
> by a force of 2mg due to gravity. Since it isn't moving initially,
> there must be an equal and opposite force: 2mg of tension in the spring.
The initial spring tension is unknown. You're assuming that the bottom
sphere is suspended from the top one. It simply says that it's suspended at
rest. Which could simply mean that the system is suspended at rest. Who
knows? Actually, I find many of these questions to be very imprecise.
Regardless, the acceleration of the system is g. And the acceleration of
all of the parts are g. Thus the string's tension should be 0.
