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Old Dec 19th 2009, 07:24 AM   #1
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Friction problem

mass of the body: 1,0 kg
A is the minimum height that the body must be abandoned to describe a circular path.

there is friction ONLY between A and C

the circular path has radius 1,0 m

1. Calculate the coefficient of kinetic friction between A and C.

2. Calculate the height from the ground that the body will leave the circular path, if left in B. (use as coefficient of friction the value of question 1.)

The image is in attachment. Please help me
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Old Dec 24th 2009, 04:02 AM   #2
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mg(3-2R)-mew mg cos 53* 3/sin 53=1/2 mgR gives mew=0.22

b)mg(2-h)- mew mg cos 53 (2/sin 53)=1/2 mg (h-R)
can be solved for h.
have u done this already?

Last edited by r.samanta; Dec 28th 2009 at 12:52 AM. Reason: ke final included
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Old Dec 25th 2009, 06:01 AM   #3
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Originally Posted by r.samanta View Post
mg(3-2R)-mew mg cos 53* 3/sin 53=0 gives mew=0.44
is this correct?
b)mg(2-h)- mew mg cos 53 (2/sin 53)=1/2 mg (h-R)
can be solved for h.
have u done this already?
to a): This is a multiple choice question and the possibilities are:
A)0,50
B)0,15
C)0,22
D)0,33
My answer is 0,22 but I'm not sure if it is correct.
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Old Dec 28th 2009, 12:51 AM   #4
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please look at my previous post again. i forgot to include final kinetic energy in the work energy theorem.
b) i think what i did is correct.did it match answers?
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Old Dec 28th 2009, 06:19 AM   #5
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Originally Posted by r.samanta View Post
please look at my previous post again. i forgot to include final kinetic energy in the work energy theorem.
b) i think what i did is correct.did it match answers?
I don't have the answers of b) =S

to b), I've done this: (in attachement)


is it correct??
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Old Dec 28th 2009, 06:34 AM   #6
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hi,
i am getting 1.4456 metres(used the equation i said earlier)
did u get that? (sorry, but your writing is appearing very small to read)
say at height h from ground, ball lose contact.ie.N=0
let theta be angle from horizontal(through centre of circle) to the point where ball lose contact.
so at that point , N(=0)+ mg cos(90- theta)=mv^2/R gives v^2=mgR sin theta
now, change in K.E= work done by gravity- work done by friction
1/2 mgRsin theta= mg(2-h)- 0.22 mg cos 53(2/sin 53)
replaced sin theta=(h-R)/R
AND GOT h=1.4456
now what is ur idea?

Last edited by r.samanta; Dec 28th 2009 at 06:41 AM.
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Old Dec 28th 2009, 09:07 AM   #7
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Originally Posted by r.samanta View Post
hi,
i am getting 1.4456 metres(used the equation i said earlier)
did u get that? (sorry, but your writing is appearing very small to read)
say at height h from ground, ball lose contact.ie.N=0
let theta be angle from horizontal(through centre of circle) to the point where ball lose contact.
so at that point , N(=0)+ mg cos(90- theta)=mv^2/R gives v^2=mgR sin theta
now, change in K.E= work done by gravity- work done by friction
1/2 mgRsin theta= mg(2-h)- 0.22 mg cos 53(2/sin 53)
replaced sin theta=(h-R)/R
AND GOT h=1.4456
now what is ur idea?
I got 1,64m but I used g=10 m s^-2 and then I used g=9,8 m s^-2 and I got 1,4456 m.

Many thanks, really!
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Old Dec 28th 2009, 09:45 AM   #8
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friend, please check the concepts again. i do not think g will affect this problem at all. i never used any value of g.it got cancelled from both sides.
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Old Dec 29th 2009, 01:01 PM   #9
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Originally Posted by r.samanta View Post
friend, please check the concepts again. i do not think g will affect this problem at all. i never used any value of g.it got cancelled from both sides.
Yes I know but I've done this exercice calculating in every step, using values so it had influenciated the calc's. I had already understood
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