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Kinematics[polar coordinate] concept problem

  • Thread starter sunumen
  • Start date
10
0
http://p14.freep.cn/p.aspx?u=v20_p14_p_0711201047589252_0.jpg [Broken]

Formula:
http://freep.cn/p.aspx?u=v20__p_0711201059233972_0.jpg [Broken]

1)I need to find out V([tex]\theta[/tex]). But I remember that r[tex]\theta[/tex]<dot>
= [tex]\omega[/tex] = V([tex]\theta[/tex])

Something seems like contradict
Where my concept wrong?
How should I deal with this problem??
I guess that the h should be included ~but why?
 
Last edited by a moderator:

rl.bhat

Homework Helper
4,433
5
The car has two velocity components. The horizontal component Vr = r*omega. It is given that h = cos(2theta) + 1. Therefore vertical component of the velocity Vh = dh/dt = -2sin(2theta)*d(theta)/dt.
The resultant of Vr and Vh gives the velocity of the car.
 
190
0
[tex]v(\theta)=\omega R(\theta)[/tex]

Distance of car to vertical shift is always changing and equals [tex]R(\theta)[/tex]
 

rl.bhat

Homework Helper
4,433
5
This variation is along the radius and hence it does not affect the theta component of the velocity.
 
190
0
But theta is on x-y plane

[tex]\omega=\frac{d\theta}{dt}[/tex]
 
Last edited:

rl.bhat

Homework Helper
4,433
5
Distance of car to vertical shift is always changing and equals Rcos(phy) This velocity does not contribute to V(theta)
 

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