How Does Velocity Change with Time in a Spiral Path?

AI Thread Summary
The discussion focuses on deriving the expression for velocity as a function of time in a spiral path using polar coordinates. The user attempts to express tangential speed and questions the correctness of their approach, specifically regarding the substitution of radial distance and differentiation. There is confusion about the dimensional correctness of the derived formula and whether the equations used are appropriate. The conversation highlights the need for clarity in applying polar coordinate transformations to velocity components. Ultimately, the user seeks validation of their calculations and understanding of the underlying physics.
roam
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Homework Statement



http://img534.imageshack.us/img534/6164/questionv.jpg

Homework Equations



ω = dθ/dt

v = dS/dt

The Attempt at a Solution



I used the second expression for the tangential speed:

v = \frac{dS}{dt} = r \frac{d \dot{\theta}}{d t} = (b-ct) \frac{d(kt)}{dt}

\therefore \ v(t) = (b-ct) k

So is this a correct expression for speed as a function of time? :confused:

So when r=0, the velocity would also be 0?
 
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in polar coordinates v2=(dr/dt)2+(rd\theta/dt)2
 
roam said:
v = \frac{dS}{dt} = r \frac{d \dot{\theta}}{d t} = (b-ct) \frac{d(kt)}{dt}
is it dimensionally correct?
 
I just substituted the rate of change of radial distance into the equation v = r.dθ/dt, and then I differentiated it. What's wrong with that? :confused:
 
I think it's dimensionally correct. Why? Did I use the wrong equations?
 
roam said:
I just substituted the rate of change of radial distance into the equation v = r.dθ/dt, and then I differentiated it. What's wrong with that? :confused:

In this case, I think this formula isn't correct.
So, <br /> \vec{r}=r\cos(\theta)\hat{x}+r\sin(\theta)\hat{y}<br /> \\<br /> \vec{v}=\frac{d\vec{r}}{dt}<br /> \\<br /> v_x=\frac{d(r\cos(\theta))}{dt}=\cos(\theta)\frac{dr}{dt}+r\frac{d(\cos\theta)}{dt}=\cos(\theta) \frac{dr}{dt}-r\sin\theta \frac{d\theta}{dt}<br /> \\<br /> v_y=\frac{d(r\sin(\theta))}{dt}=\sin(\theta)\frac{dr}{dt}+r\cos\theta\frac{d\theta}{dt}<br /> \\<br /> v^2=(v_x)^2+(v_y)^2=(\frac{dr}{dt})^2+(r\frac{d \theta}{dt})^2<br /> <br />
 
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