In non-uniform circular motion, is the net acceleration still pointing to the centre? Stuck on a v = kt problem

7 answers & replies

• [+][–]Arjun Nair12thSample· 19d ago

  Two components, both perpendicular:

  • Centripetal (radial, towards the centre): $a_c = v^2/R$. It changes the direction of velocity.
  • Tangential (along the velocity): $a_t = dv/dt$. It changes the speed.

  For $v=kt$: $a_t = k$ and $a_c = k^2t^2/R$. They are perpendicular, so $a = \sqrt{a_t^2 + a_c^2} = \sqrt{k^2 + \frac{k^4t^4}{R^2}} = k\sqrt{1+\frac{k^2t^4}{R^2}}$ Angle $\phi$ with the radius: $\tan\phi = \dfrac{a_t}{a_c} = \dfrac{R}{kt^2}$. At $t=0$ this is 90° (all tangential) and as $t$ grows the vector swings towards the centre.

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  • [+][–]Ishita GoyalDropperSample· 19d agoAuthor

    Template box filled. Added "a_t changes speed, a_c changes direction" in red. Thanks all, especially for settling the Rα thing.

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• [+][–]Tushar PawarDropperSample· 19d ago

  Your second component is $a_t = dv/dt$, which is not $v^2/R$. If someone writes $a_t = v^2/R$ they've mixed it up with the radial one. Also don't differentiate $v^2/R$ with respect to time and call that tangential.

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  • [+][–]Kabir Joshi11thSample· 19d ago

    Hmm, I thought tangential acceleration is $R\alpha$ where $\alpha$ is angular acceleration, so isn't that a different thing than $dv/dt$?

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    • [+][–]Aman YadavDropperSample· 19d ago

      Same thing. $v = R\omega$, so $dv/dt = R\,d\omega/dt = R\alpha$. Two ways to write the same $a_t$. Use whichever the question gives you.

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• [+][–]Neel Chatterjee12thSample· 19d ago

  Vertical circle with a ball on a string. Gravity has a tangential component $mg\sin\theta$ except at two points. So the speed changes all the time, and the net acceleration is not directed to the centre in general.

  Exactly at the top and bottom, gravity is purely radial (along the string line), so $a_t = 0$ there and the net acceleration points to the centre. Nice test question: "at what points is the acceleration directed along the radius". Answer: top and bottom (and nowhere else for non-uniform motion unless the speed is constant).

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• [+][–]Vivek Mishra12thSample· 19d ago

  Put numbers in your template box so you can feel it. Take $k=2\ \text{m/s}^2$, $R=2$ m, $t=1$ s. Then $v=2$ m/s, $a_t=k=2\ \text{m/s}^2$, $a_c=v^2/R=4/2=2\ \text{m/s}^2$. Both equal, so net $a=2\sqrt2\approx2.83\ \text{m/s}^2$ and the angle with the radius is $45^\circ$. That matches $\tan\phi=R/(kt^2)=1$. If $t$ is smaller, the tangential part dominates, if $t$ is larger the radial part does.

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