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experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1350734665748:dw
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Third case: T = g?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
well ... its g N
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
first case??
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Wait. N = unit?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yes ... tension always has unit of force ... except for surface tension.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
First case: Σ F = mg mg  T = mg T = 0
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Second case: Σ F = ma mg  T = ma T = m (ga) = m(g5) Unit: N for first and second case.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
But..... but.... but.... the reel.... Shouldn't I take it into consideration??
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
you get acceleration after considering the rotational motion of reel. it shouldn't make any difference.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
sorry .. mg*R  TR = torque being applied on reel ... that makes it go round.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
mg*R  TR < this part creates motion.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
acc. of the reel = acc. of the mass?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
no ... the reel does not have linear acceleration. It's rotating ... it has angular acceleration.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
tangential acc. of the reel = acc. of teh mass?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yes ... maybe ... worth checking it out.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
If we don't have the acc. of the reel, how can we suppose to find torque, so that we can find tension!?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
well .. we have to find the acceleration reel first.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
or acceleration of the mass falling down first.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
But.. how...
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
solve for acceleration of the using it.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Wow! We're doing it backward!
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
(c) > (b) > (a) :O
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
well ...
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
That's not a problem :P
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
the other way ... "tangential acc. of the reel = acc. of teh mass" ... at the circumference of reel ... they go foot to foot.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yep .. this is also possible.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
For the energy one, can we really use that way?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
sure ... the torque will always be constant ... hence there will be constant acceleration.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
I'm sorry, but how do I know if the torque is constant?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
mgh?! the PE of the reel or..?!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
mass .. hold on ... there seems to be discrepancy with second method.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1350737767080:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1350737955528:dw
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Hmm.. So, you're just considering the mass m ..?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1350738106593:dw
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
dw:1350738529141:dw
 2 years ago