experimentX
  • experimentX
Q.
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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experimentX
  • experimentX
|dw:1350734665748:dw|
Callisto
  • Callisto
Third case: T = g?
experimentX
  • experimentX
well ... its g N

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experimentX
  • experimentX
first case??
Callisto
  • Callisto
Wait. N = unit?
experimentX
  • experimentX
yes ... tension always has unit of force ... except for surface tension.
Callisto
  • Callisto
First case: Σ F = mg mg - T = mg T = 0
Callisto
  • Callisto
Second case: Σ F = ma mg - T = ma T = m (g-a) = m(g-5) Unit: N for first and second case.
Callisto
  • Callisto
But..... but.... but.... the reel.... Shouldn't I take it into consideration??
experimentX
  • experimentX
you get acceleration after considering the rotational motion of reel. it shouldn't make any difference.
experimentX
  • experimentX
in other words you have already taken care of rotational motion of reel. still if you need to verity then ... T*R = torque being applied on reel.
experimentX
  • experimentX
sorry .. mg*R - TR = torque being applied on reel ... that makes it go round.
experimentX
  • experimentX
mg*R - TR <--- this part creates motion.
Callisto
  • Callisto
acc. of the reel = acc. of the mass?
experimentX
  • experimentX
no ... the reel does not have linear acceleration. It's rotating ... it has angular acceleration.
Callisto
  • Callisto
tangential acc. of the reel = acc. of teh mass?
experimentX
  • experimentX
yes ... maybe ... worth checking it out.
Callisto
  • Callisto
If we don't have the acc. of the reel, how can we suppose to find torque, so that we can find tension!?
experimentX
  • experimentX
well .. we have to find the acceleration reel first.
experimentX
  • experimentX
or acceleration of the mass falling down first.
Callisto
  • Callisto
But.. how...
experimentX
  • experimentX
using conservation of energy \[ {1 \over 2} mv^2 + {1 \over 2}I \omega^2 = mgh \\ \omega r = v,\] solving those two will give you final velocity, you know distance 'd', and initial velocity.
experimentX
  • experimentX
solve for acceleration of the using it.
Callisto
  • Callisto
Wow! We're doing it backward!
Callisto
  • Callisto
(c) -> (b) -> (a) :O
experimentX
  • experimentX
well ...
Callisto
  • Callisto
That's not a problem :P
experimentX
  • experimentX
the other way ... "tangential acc. of the reel = acc. of teh mass" ... at the circumference of reel ... they go foot to foot.
Callisto
  • Callisto
Yes. Then we can find the angular acc. Using integration to find the moment of inertia, and then use it to find the torque. Is it possible?
experimentX
  • experimentX
yep .. this is also possible.
Callisto
  • Callisto
For the energy one, can we really use that way?
experimentX
  • experimentX
sure ... the torque will always be constant ... hence there will be constant acceleration.
Callisto
  • Callisto
I'm sorry, but how do I know if the torque is constant?
experimentX
  • experimentX
the torque depends on two parameters \( r \times F\) , F is part of mgh while r is radius of wheel. both are not changing ... neither is their angle between them.
Callisto
  • Callisto
mgh?! the PE of the reel or..?!
Callisto
  • Callisto
hmm?
experimentX
  • experimentX
mass .. hold on ... there seems to be discrepancy with second method.
experimentX
  • experimentX
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experimentX
  • experimentX
|dw:1350737955528:dw|
Callisto
  • Callisto
Hmm.. So, you're just considering the mass m ..?
experimentX
  • experimentX
|dw:1350738106593:dw|
Callisto
  • Callisto
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