My Math Forum How can I relate the arc length to the acceleration of an object with rotation?

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 November 5th, 2019, 02:10 AM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus How can I relate the arc length to the acceleration of an object with rotation? The problem is as follows: A spheric strawberry candy is spinning inside a centrifuge in a confectionery. Find the magnitude of the acceleration in $\frac{m}{s^{2}}$ of the candy after $4\,s$ from the beginning of its rotation. It is known that the arc length follows the function $s(t)=1+t^3$ where $t$ is the elapsed time. It is also known that after $2\,s$ from its rotation the normal (centripetal) acceleration is $2\,\frac{m}{s^{2}}$. The given alternatives in my book are as follows: $\begin{array}{ll} 1.&40\,\frac{m}{s^{2}}\\ 2.&35\,\frac{m}{s^{2}}\\ 3.&30\,\frac{m}{s^{2}}\\ 4.&45\,\frac{m}{s^{2}}\\ 5.&50\,\frac{m}{s^{2}}\\ \end{array}$ The only think which I can recall here is that the acceleration is the second derivative of the length. Since the function over time is given then that would be the tantential acceleration. Since $s(t)=1+t^3$ Then $s'(t)=3t^2$ and $s''(t)=6t$ So by evaluating this function on t=4 gives me the tangential acceleration at that given instant. $s''(4)=6(4)=24$ But then what?. I can relate that the centripetal acceleration (which is what is being asked) is given by: $a_{c}=\omega^2 r$ However the angular speed to be used here is that attained after $4$ seconds have elapsed. But the problem is to find the radius. But we don't know that: But we do know what is the value of the arc length. $s(2)=1+2^3=9\,m$ and the centripetal acceleration for that given time: $a_{c}=\omega^2 r = 2$ So what we need is the angular speed at $2\,s$ as a function of the radius. Since what we need here is the angular acceleration, we can relate it with the tangential acceleration by the equation: $a_{t}=6t=6\times 2 = 12$ $\alpha=\frac{a_{t}}{r}=\frac{12}{r}$ Therefore we have the angular acceleration in terms of the radius. Thus: $\omega_{f}=\omega_{o}+\alpha t$ $\omega_{f}=0+\frac{12}{r}\times 2 = \frac{24}{r}$ Then we have the angular speed as a function of $r$. This can be replaced in the given value for the centripetal acceleration. $a_{c}=\omega^2 r = 2$ $\left(\frac{24}{r}\right)^2r = 2$ Therefore: $r=288\,m$ Then using this value we can compute the centripetal acceleration: But we do need the angular acceleration for $t=4$: $\alpha=\frac{a_{t}}{r}=\frac{24}{288}= \frac{1}{12}$ I noted that this is different than the angular acceleration at $t=2$ $\alpha=\frac{12}{288}= \frac{1}{24}$ Now arises the question. What sort of "acceleration" is the problem asking? Perhaps is the magnitude of the total acceleration or the norm that is happening because the tangential acceleration and the centripetal acceleration? If so that would be: $a=\sqrt{\left(\frac{1}{12}\right)^{2}+(24)^{2}}$ Then this would become into: $a\approx 24$ But this doesn't appear in any of the alternatives given. Can somebody help me here?. What's exactly the part where I got lost?
 November 7th, 2019, 12:49 AM #2 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus Can anyone take a look into this question please? Did I get something wrong? Last edited by skipjack; November 8th, 2019 at 04:50 AM.
 November 7th, 2019, 05:14 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 $\displaystyle v_T(t) = s’(t) = 3t^2 \implies a_c(t) = \frac{9t^4}{r}$ $\displaystyle a_c(2)= \frac{9 \cdot 16}{r} = 2 \implies r = 72 \, m$ $\displaystyle a_c(4) = \frac{9 \cdot 4^4}{72} = 32 \, m/s^2$ $s = r \cdot \theta \implies s(t) = r \cdot \theta(t) = 1 + t^3$ $\displaystyle \frac{d^2\theta}{dt^2} = \alpha(t) = \frac{6t}{r}$ $a_T(4) = r \cdot \alpha(4) = 24 \, m/s^2$ $a(4) = \sqrt{a_c(4)^2 + a_T(4)^2} = \sqrt{(4 \cdot 8 )^2 + (3 \cdot 8 )^2} = 5 \cdot 8 = 40 \, m/s^2$ Thanks from Chemist116 Last edited by skipjack; November 8th, 2019 at 04:48 AM.
November 8th, 2019, 02:41 AM   #4
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Joined: Jun 2017
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Math Focus: Calculus

Quote:
 Originally Posted by skeeter $\displaystyle v_T(t) = s’(t) = 3t^2 \implies a_c(t) = \frac{9t^4}{r}$ $\displaystyle a_c(2)= \frac{9 \cdot 16}{r} = 2 \implies r = 72 \, m$ $\displaystyle a_c(4) = \frac{9 \cdot 4^4}{72} = 32 \, m/s^2$ $s = r \cdot \theta \implies s(t) = r \cdot \theta(t) = 1 + t^3$ $\displaystyle \frac{d^2\theta}{dt^2} = \alpha(t) = \frac{6t}{r}$ $a_T(4) = r \cdot \alpha(4) = 24 \, m/s^2$ $a(4) = \sqrt{a_c(4)^2 + a_T(4)^2} = \sqrt{(4 \cdot 8 )^2 + (3 \cdot 8 )^2} = 5 \cdot 8 = 40 \, m/s^2$
From the looks of it, it seems that they intended to ask the total acceleration, as I see you used the norm. I feel this was a conceptual gap from my end. I totally overlooked that the tangential speed could be obtained from the first derivative of the arc length. But I wonder whether the procedure that I intended to do was too way off from target. Why I got the radius bigger than you? Can you perhaps look into my steps and tell me what exactly I did wrong so I don't fall into the same mistake in the future?

Last edited by skipjack; November 8th, 2019 at 04:49 AM.

 November 8th, 2019, 09:28 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 note that the equation $\omega = \omega_0 + \alpha t$ only works if $\alpha$ is a constant ... in this case, $\alpha = \dfrac{a_T}{r} = \dfrac{6t}{r}$ is variable. $\displaystyle \omega(t) = \omega_0 + \int_0^t \dfrac{6x}{r} \, dx$ $\displaystyle \omega(t) = \omega_0 + \bigg[\dfrac{3x^2}{r} \bigg]_0^t$ $\omega(t) = 0 + \dfrac{3t^2}{r}$ $\omega(2) = \dfrac{12}{r}$ $a_c(2) = r \cdot [\omega(2)]^2 = r \cdot \dfrac{12^2}{r^2} = \dfrac{144}{r}$ $2 = \dfrac{144}{r} \implies r = 72 \, m$

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