Centripetal Force & Gravitation

  • #1
it's on centripetal force and gravitation problems..
wonderin if anyone can provide me answers such that i can compare mine
here's the assignment
http://www.geocities.com/cheemaharmeet/assignment.jpg

mine are {
1. 8N;
2. 1.97 hr;
3. 7.9E3 m/s
4. 382.26 m, 5.49E3 N
5. 8E-10 N
6. 2.93E41 kg}
I would really thankful, is someone can provide me answers..
and if u can do atleast.. please try #5.. i am confused over that.because my book also have answer to that question, and mine one's different from that
 
  • #2
I don't get the same answer for #4. Can you show more of how you got it? Perhaps I made an error...

Regarding #5, what is the answer the book gives? I think I know why you got a different answer.
 
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  • #3
I get the same answers as you on all questions except #4. Can you show more of how you got it? Perhaps I made an error...

thnx a lot for assuring me..
and for #4

N-W = centripetal force
m.(normal a)-mg = m(v^2/r)

but normal a=7g

so
(v^2/r)=7g-g

so r=(v^2/g)
=382.23 m

and
N = 7 x 9.81 x 80 kg


I think i am wrong, perhaps because in the question it says "acceleration of the pilot"..
so that means net acceleration?
 
  • #4
Note that I edited my first post about question #5. Can you tell me what the book's answer is?
 
  • #5
Note that I edited my first post about question #5. Can you tell me what the book's answer is?

6.8 N
umm.. i think this isn't possible though
 
  • #6
(v^2/r)=7g-g

so r=(v^2/g)
=382.23 m

Even if you do it this way, shouldn't it be 6g on the bottom (7g-g=6g)?

The wording is bit confusing in the question. I would think they are just asking for the acceleration, so then r = (v^2)/7g.
To find the force of the seat on the pilot, you just sum the forces up, so N - mg = mv^2/r. That's the approach I would take, but you decide what you want to do.
 
  • #7
oops, that was a typo
umm.. yea, i think your way''s right
 
  • #8
6.8 N
umm.. i think this isn't possible though

Is that supposed to be 6.8x10^-10 N though? Gravitational force is a vector. So for the third mass (4kg), you need to break up the force into its x and y components. Then do a sum of the forces in each direction. Then find the resultant of those two. Does that make sense? If you do that, you should get the answer the book gives.
 
  • #9
Yes 6.8x10^-10 N is the correct answer for 5/.

Regards.

Nacer.
 

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