Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Q.

OCW Scholar - Single Variable Calculus
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

|dw:1350734665748:dw|
Third case: T = g?
well ... its g N

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

first case??
Wait. N = unit?
yes ... tension always has unit of force ... except for surface tension.
First case: Σ F = mg mg - T = mg T = 0
Second case: Σ F = ma mg - T = ma T = m (g-a) = m(g-5) Unit: N for first and second case.
But..... but.... but.... the reel.... Shouldn't I take it into consideration??
you get acceleration after considering the rotational motion of reel. it shouldn't make any difference.
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.
sorry .. mg*R - TR = torque being applied on reel ... that makes it go round.
mg*R - TR <--- this part creates motion.
acc. of the reel = acc. of the mass?
no ... the reel does not have linear acceleration. It's rotating ... it has angular acceleration.
tangential acc. of the reel = acc. of teh mass?
yes ... maybe ... worth checking it out.
If we don't have the acc. of the reel, how can we suppose to find torque, so that we can find tension!?
well .. we have to find the acceleration reel first.
or acceleration of the mass falling down first.
But.. how...
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.
solve for acceleration of the using it.
Wow! We're doing it backward!
(c) -> (b) -> (a) :O
well ...
That's not a problem :P
the other way ... "tangential acc. of the reel = acc. of teh mass" ... at the circumference of reel ... they go foot to foot.
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?
yep .. this is also possible.
For the energy one, can we really use that way?
sure ... the torque will always be constant ... hence there will be constant acceleration.
I'm sorry, but how do I know if the torque is constant?
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.
mgh?! the PE of the reel or..?!
hmm?
mass .. hold on ... there seems to be discrepancy with second method.
|dw:1350737767080:dw|
|dw:1350737955528:dw|
Hmm.. So, you're just considering the mass m ..?
|dw:1350738106593:dw|
|dw:1350738529141:dw|