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My lecturer states that $a_r=\dfrac{v_t^2}{r}=\omega^2r$ where $v_t$ is tangential velocity, he also wrote that this is derived the same way that radial acceleration is derived in uniform circular motion.

I know the derivation for uniform circular motion but I simply can't see how the derivation can be the same for non uniform motion. (the derivation I know is the graphical one where you manipulate the infinitesimal velocity and angular position.)

Qmechanic
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Luca Ion
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1 Answers1

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For a non-uniform circular motion your graphical demonstration remains valid. Simply, $\frac{d\theta}{dt}=\omega$ is no longer constant in time. You could write $\omega(t)$ to make it explicitly time-dependent.

user8736288
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  • But, if you use the graphical method, then when you do v(final)-v(initial) after dt time , the vector dv is not in same direction like the radial acceleration. Vector dv has direction =a(tangential)+a(radial) isn't it ? – Luca Ion Oct 04 '19 at 08:08
  • Yes you're right! I have probably overlooked the difficulty of deriving it from a graphical method. You should probably follow a more general demonstration in this case, as suggested by G. Smith. – user8736288 Oct 04 '19 at 12:27