Titled reference frame, N2L with position and velocity

In summary: sorry, the equation for y is?' y= tan(\phi)x= (sin(\phi)/cos(\phi))x ' is because when you graphed the equation for x, it looks like a parabola.
  • #36
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]
 
Last edited:
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  • #37
i have a feeling this is wrong for dy...
 
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  • #38
Oblio said:
i have a feeling this is wrong for dy...

Looks right to me. So now you want to find the range...
 
  • #39
range?
 
  • #40
Oblio said:
range?

the question asks for the range...
 
  • #41
which is really just the displacement up the plane?
 
  • #42
Oblio said:
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...
 
  • #43
is this the same kinematics equation?
 
  • #44
Oblio said:
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.
 
  • #45
solve for... vo? and insert into dx?
 
  • #46
Oblio said:
solve for... vo? and insert into dx?

no, you want to eliminate t... so solve for t and substitute into dx.
 
  • #47
without simplifying yet..

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

ya?
 
  • #48
Oblio said:
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?
 
  • #49
t = vosin(theta) / (-1/2)gsin(phi)
 
  • #50
Oblio said:
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.
 
  • #51
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!
 
  • #52
Oblio said:
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!

cool. plugging that into dx should work.
 
  • #53
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)
 
  • #54
Oblio said:
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.
 
  • #55
oops.
man I am bad..
 
  • #56
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
 
  • #57
Oblio said:
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...
 
  • #58
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)
 
  • #59
Oblio said:
oops again.

so I am 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...
 
  • #60
will trig identities apply since theta and phi are present?
 
  • #61
im confused how one would ever get, for example : cos (theta + phi) through simplifying
 
  • #62
Oblio said:
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)...
 
  • #63
i found that cos (a+b) = cosacosb +/- sinasinb... but i don't have that relationship anywhere
 
  • #64
Oblio said:
i found that cos (a+b) = cosacosb +/- sinasinb... but i don't 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.
 
  • #65
k I am at
dx= vo^2sin(theta)*(cos(theta)-sin(phi)) / (1/2)cos(phi)^2(sin(theta))
 
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  • #66
Oblio said:
k I am 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...
 
  • #67
i can't factor out a vo^2?
 
  • #68
Oblio said:
i can't 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
 
  • #69
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
 
  • #70
Oblio said:
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.
 

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