In article "Parse Tree" <parsetree at hotmail.com> writes:
<
<"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.
<
This suddenly grows more interesting. I've been reading "spring"; it
in fact says "string".
To be pedantic, in the real world, my argument would still apply --
a string is over short ranges in fact a spring, one with an extremely large
spring constant. But that's obviously not the intent of the question.
I mis-read it.
<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.
<
If you assume an infinitely strong string, then you are correct. Otherwise
they will indeed move, unless you've stretched the string inelastically.
However, I'm being picky, and you're on to the intention of the question,
I think.
<> <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.
Assuming an infinitely strong string -- one whose relaxation is zero -- then
you are correct.
-- cary
