# Kinematics[polar coordinate] concept problem

1. Nov 19, 2007

### sunumen

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($$\theta$$). But I remember that r$$\theta$$<dot>
= $$\omega$$ = V($$\theta$$)

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: May 3, 2017
2. Nov 20, 2007

### rl.bhat

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.

3. Nov 20, 2007

### azatkgz

$$v(\theta)=\omega R(\theta)$$

Distance of car to vertical shift is always changing and equals $$R(\theta)$$

4. Nov 20, 2007

### rl.bhat

This variation is along the radius and hence it does not affect the theta component of the velocity.

5. Nov 20, 2007

### azatkgz

But theta is on x-y plane

$$\omega=\frac{d\theta}{dt}$$

Last edited: Nov 20, 2007
6. Nov 21, 2007

### rl.bhat

Distance of car to vertical shift is always changing and equals Rcos(phy) This velocity does not contribute to V(theta)