Question

a particle describes an equiangular spiral r=aeki power theata in such a manner that it's acceleration has no radial component. prove that it's angular vvilocity is constant and that the magnitude of the vilocity and acceleration is each proportional to r.
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Answer 1

Explanation:

I use the full equation for the logarithmic spiral with parameter b.

To iterate, we are given r = a e^(bθ) with a, b CONSTANTS.

Then the linear velocity is v = dr/dθ = ab e^(bθ) = b r

Linear acceleration = a = dv/dθ = d(b

Answer 2

Concept:

An individual parameter spiral family is referred to as an equiangular spiral. It is described as a curve that radially intersects all other lines at a fixed angle. It is also known as logistique, Bernoulli spiral, and the logarithmic spiral.

Given:

The equation of the curve is $r =ae^{\theta}$.

Find:

To prove that the acceleration is also proportional to r.

Solution:

The equation of the curve is $r =ae^{\theta}$.

$\frac{dr}{dt} = ae^{\theta} = r\frac{dr}{dt}..(1) \\\frac{d^{2}r }{dt^{2} } = \frac{dr}{dt} .\frac{d\theta}{dt} + r\frac{d^{2}r }{dt^{2} } \\\frac{d^{2}r }{dt^{2} } = r(\frac{d\theta}{dt} )^{2} + r\frac{d^{2} \theta}{dt^{2} }..(2)$

The particle's acceleration has no radial component.

Radial acceleration = $\frac{d^{2}r }{dt^{2} } - r (\frac{d\theta}{dt} )^{2} = 0$

$\frac{d^{2}r }{dt^{2} } = r (\frac{d\theta}{dt} )^{2}..(3)$

From (2) and (3), we get:

$r (\frac{d\theta}{dt} )^{2} + r\frac{d^{2}r }{dt^{2} } = r(\frac{d\theta}{dt} )^{2}\\r\frac{d^{2} \theta}{dt^{2} } =0\\\frac{d^{2}\theta }{dt^{2} } =0\\\frac{d^{2} \theta}{dt^{2} } =0 or \frac{d\theta}{dt } =constant$

The angular velocity is constant.

Now, from (1) $\frac{dr}{dt} = r \omega$

We have to also solve the magnitude of the velocity

V = $\sqrt{(\frac{dr}{dt} )^{2} + (r\frac{d\theta}{dt} )^{2}$

V = $\sqrt{r^{2}\omega^{2} +r^{2}\omega^{2} } = r\omega\sqrt{2}$

Again transverse acceleration = 2ω(rω) = 2ω²r

Hence, we have proved that acceleration is also proportional to r.

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