Titled reference frame, N2L with position and velocity


by Oblio
Tags: frame, position, reference, titled, velocity
Oblio
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#55
Sep14-07, 04:30 PM
P: 397
oops.
man im bad..
Oblio
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#56
Sep14-07, 04:33 PM
P: 397
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2
learningphysics
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#57
Sep14-07, 04:35 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))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2
careful the 1/2 and g are squared...
Oblio
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#58
Sep14-07, 04:38 PM
P: 397
oops again.

so im left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)
learningphysics
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#59
Sep14-07, 04:43 PM
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Quote Quote by Oblio View Post
oops again.

so im left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)
Looks good... try to simplify and use a trig identity to get it to look like the formula they gave for the range...
Oblio
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#60
Sep14-07, 04:45 PM
P: 397
will trig identities apply since theta and phi are present?
Oblio
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#61
Sep14-07, 04:46 PM
P: 397
im confused how one would ever get, for example : cos (theta + phi) through simplifying
learningphysics
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#62
Sep14-07, 05:04 PM
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Quote Quote by Oblio View Post
im confused how one would ever get, for example : cos (theta + phi) through simplifying
try a little factoring of your equation also... look up the identity for cos(A+B)...
Oblio
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#63
Sep14-07, 05:06 PM
P: 397
i found that cos (a+b) = cosacosb +/- sinasinb... but i dont have that relationship anywhere
learningphysics
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#64
Sep14-07, 05:09 PM
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Quote Quote by Oblio View Post
i found that cos (a+b) = cosacosb +/- sinasinb... but i dont have that relationship anywhere
first write everything over 1 denominator... then compare what you have to the formula you need to get... a little factoring will give you the answer.
Oblio
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#65
Sep14-07, 05:14 PM
P: 397
k im at
dx= vo^2sin(theta)*(cos(theta)-sin(phi)) / (1/2)cos(phi)^2(sin(theta))
learningphysics
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#66
Sep14-07, 05:21 PM
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Quote Quote by Oblio View Post
k im at
dx= vo^2sin(theta)*(cos(theta)-sin(phi)) / (1/2)cos(phi)^2(sin(theta))
factoring out the sin(theta) was correct... but you made a mistake somewhere...
Oblio
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#67
Sep14-07, 05:22 PM
P: 397
i cant factor out a vo^2?
learningphysics
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#68
Sep14-07, 05:23 PM
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Quote Quote by Oblio View Post
i cant factor out a vo^2?
yes you can... I was referring to the sin's cos's... check your work... your denominator should only have [cos(phi)]^2
Oblio
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#69
Sep14-07, 05:25 PM
P: 397
i edited that in by mistake. i meant to put that in the numerator.

dx= vo^2sin(theta)*(cos(theta)-sin(phi)sin(theta) / (1/2)gcos(phi)^2
learningphysics
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#70
Sep14-07, 05:27 PM
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Quote Quote by Oblio View Post
i edited that in by mistake. i meant to put that in the numerator.

dx= vo^2sin(theta)*(cos(theta)-sin(phi)sin(theta) / (1/2)gcos(phi)^2
almost there... should be cos(theta)cos(phi) in the numerator... I think you forgot to multiply the top by cos(phi) when putting everything over 1 denominator.
Oblio
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#71
Sep14-07, 05:31 PM
P: 397
ya, you mean i didnt multiply it by (1/2)gcos(phi)^2?
Oblio
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#72
Sep14-07, 05:50 PM
P: 397
ignore that last one


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