Titled reference frame, N2L with position and velocity


by Oblio
Tags: frame, position, reference, titled, velocity
Oblio
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#37
Sep14-07, 03:28 PM
P: 397
i have a feeling this is wrong for dy....
learningphysics
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#38
Sep14-07, 03:32 PM
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Quote Quote by Oblio View Post
i have a feeling this is wrong for dy....
Looks right to me. So now you want to find the range...
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#39
Sep14-07, 03:36 PM
P: 397
range?
learningphysics
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#40
Sep14-07, 03:37 PM
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Quote Quote by Oblio View Post
range?
the question asks for the range...
Oblio
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#41
Sep14-07, 03:40 PM
P: 397
which is really just the displacement up the plane?
learningphysics
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#42
Sep14-07, 03:43 PM
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Quote Quote by Oblio View Post
which is really just the displacement up the plane?
yes, maximum displacement... the object is thrown at an angle, then hits the incline eventually... the displacement up the plane when it hits is the range...
Oblio
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#43
Sep14-07, 03:45 PM
P: 397
is this the same kinematics equation?
learningphysics
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#44
Sep14-07, 03:49 PM
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Quote Quote by Oblio View Post
dx = vocos[tex]\vartheta[/tex]t + (1/2)(-g)t^2sin[tex]\phi[/tex]

dy = vosin[tex]\vartheta[/tex]t + (1/2)(-g)t^2cos[tex]\phi[/tex]
use these equations to find dx when dy = 0... that gives the range.
Oblio
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#45
Sep14-07, 03:52 PM
P: 397
solve for...... vo? and insert into dx?
learningphysics
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#46
Sep14-07, 03:55 PM
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Quote Quote by Oblio View Post
solve for...... vo? and insert into dx?
no, you want to eliminate t... so solve for t and substitute into dx.
Oblio
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#47
Sep14-07, 04:02 PM
P: 397
without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?
learningphysics
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#48
Sep14-07, 04:08 PM
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Quote Quote by Oblio View Post
without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?
hmm... close but some mistakes... can you post what you got for t?
Oblio
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#49
Sep14-07, 04:09 PM
P: 397
t = vosin(theta) / (-1/2)gsin(phi)
learningphysics
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#50
Sep14-07, 04:17 PM
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Quote Quote by Oblio View Post
t = vosin(theta) / (-1/2)gsin(phi)
setting dy = 0, and solving for t will give you something different... it should be cos(phi) and the - shouldn't be there.
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#51
Sep14-07, 04:19 PM
P: 397
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!
learningphysics
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#52
Sep14-07, 04:21 PM
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Quote Quote by Oblio View Post
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!
cool. plugging that into dx should work.
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#53
Sep14-07, 04:24 PM
P: 397
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)
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#54
Sep14-07, 04:29 PM
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Quote Quote by Oblio View Post
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)
you missed t^2... didn't square t.


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