non uniform circular motion?

Question

particle is moving in circular motion such that its normal and tangential accelerations are equal. find the velocity of the particle after moving a distance s

Answer 1

v0*exp(+/- s/ρ)

an = ρ*ω^2
at = +/- ρ*dω/dt
Because accelation is a vector, the magnitudes of an and at are considered equal in this problem which does not specify whether the particle is speeding up or slowing down.

So an = at means
dω/dt = +/- ω^2
dω/ω^2 = +/- dt
Integrating
-1/ω + 1/ω0 = +/- t

ω = ω0 / (1 +/- ω0 * t)
Integrate to find θ
θ = +/- ln|1 +/- ω0 * t|

s = ρ * θ
So +/- ln|1 +/- ω0 * t| = s/ρ
1 +/- ω0 * t = exp(-/+ s/ρ)

ω = ω0 / exp(-/+ s/ρ)
v = ρ ω = v0 * exp(+/- s/ρ)

Answer 2

Given
a (tangential) = a (radial) = v^2 / r
{Magnitudes alone can be equal}

dv/dt = v^2 / r

dv / v^2 = dt / r

Integrating both sides

- 1/v = t /r + C

Let at time t = 0, v = u. C = - 1/u

- 1/v = t /r - 1/u

- 1/v = t /r- 1/u

v = ur / [r –ut]
ds/dt = ur / [r –ut]

ds ={ ur / [r –ut]} dt.
Integrating both sides
S = - r ln [r –ut]} + C To elimlinate t,

usung v = ur / [r –ut] , r- ut = v/[ur]

S = - r ln v/[ur] + C
S= C - r ln (v / [ur])
When v = u , initial condition, let S = 0

0= C - r ln ( 1/ r]

C = ln ( 1/ r ^r ]

S = ln ( 1/ r ^r ] - r ln (v / [ur])

S = = ln {u ^r r^(r-r) / v^r }

e^S / r = (u / v}

v = u r / e^S

Answer 3

an object covering circular path in unequal interval of time

Answer 4

depends on initial speed

Answer 5

rg