small centripetal acceleration question (1 Viewer)

jathu123 · Mar 13, 2016

So I'm pretty confused as in how to approach this question:
A car of mass 450 kg travelling North at 50.0 km/h changes direction and is travelling at 50.0 km/h due east 20.0 s later, while travelling around a corner.
a) Find the change in velocity

My answer for this:

$since change in velocity is $ \Delta v=v_{f}-v_{i}\\ =v_{f}+(-v_{i}) \\ = \sqrt{50^2 + 50^2} \\ =70.7 $ km/h towards the centre$

But the next question is what confuses me:
b) find the centripetal acceleration

$$ My working out: $ a_{c}=\frac{v^2}{R} $ where tangential velocity is $ \frac{50}{3.6}ms^{-1} $ but the Radius is unknown. So I used the formula $ v=\frac{d}{t} = \frac{2\pi R}{T}. $ Since the car took 20 seconds to travel 1/4th of the circumference, the period is 80 seconds. Now by solving for R, I got $ 173.8 m. $ Plugging it back into the acceleration formula, I got $ a_{c} =1.09ms^{-2}. $ But by looking back at part a, I now think they want me to find the acceleration by doing $ \frac{\Delta v}{\Delta t}=\frac{19.64}{20}=0.98ms^{-2}.$

I know the answers may look similar, but I did not do any rounding up. So which is the correct way? I think the first one is, but by looking at the order of these questions, it really confuses me (in fact, the last part is the one on finding the radius, which I already did to calculate part b lol). Thanks

Green Yoda · Mar 13, 2016

latex on point jathu

jathu123 · Mar 13, 2016

Rathin said:
latex on point jathu

haha legit took me an hour to type it up

InteGrand · Mar 13, 2016

jathu123 said:
So I'm pretty confused as in how to approach this question:
A car of mass 450 kg travelling North at 50.0 km/h changes direction and is travelling at 50.0 km/h due east 20.0 s later, while travelling around a corner.
a) Find the change in velocity

My answer for this:

$since change in velocity is $ \Delta v=v_{f}-v_{i}\\ =v_{f}+(-v_{i}) \\ = \sqrt{50^2 + 50^2} \\ =70.7 $ km/h towards the centre$

But the next question is what confuses me:
b) find the centripetal acceleration

$$ My working out: $ a_{c}=\frac{v^2}{R} $ where tangential velocity is $ \frac{50}{3.6}ms^{-1} $ but the Radius is unknown. So I used the formula $ v=\frac{d}{t} = \frac{2\pi R}{T}. $ Since the car took 20 seconds to travel 1/4th of the circumference, the period is 80 seconds. Now by solving for R, I got $ 173.8 m. $ Plugging it back into the acceleration formula, I got $ a_{c} =1.09ms^{-2}. $ But by looking back at part a, I now think they want me to find the acceleration by doing $ \frac{\Delta v}{\Delta t}=\frac{19.64}{20}=0.98ms^{-2}.$

I know the answers may look similar, but I did not do any rounding up. So which is the correct way? I think the first one is, but by looking at the order of these questions, it really confuses me (in fact, the last part is the one on finding the radius, which I already did to calculate part b lol). Thanks

$\noindent Remember, the formula $a_c = \frac{v^2}{r}$ gives us the magnitude of the centripetal acceleration vector, \emph{which is by definition the limit of the average acceleration vector as the time increment is made very small}, i.e. $\overrightarrow{a_c}= \lim_{\Delta t\to 0}\frac{\Delta \overrightarrow{v}}{\Delta t}=\overrightarrow{v}^\prime (t)$. So the answer obtained from $\frac{\Delta v}{\Delta t}$ is the \emph{average acceleration} over the 20 seconds, which is not the same as the centripetal acceleration. As you made the $\Delta t$ smaller and smaller, it would converge to the value you got for $a_c$. So technically your original answer is correct (the $1.09$ one) if they asked for the magnitude of the centripetal acceleration. If they asked for the magnitude of the $\emph{average acceleration}$, then the latter one would be right.$

jathu123 · Mar 13, 2016

InteGrand said:
$\noindent Remember, the formula $a_c = \frac{v^2}{r}$ gives us the centripetal acceleration, \emph{which is by definition the limit of the average acceleration as the time increment is made very small}, i.e. $a_c = \lim_{t\to 0}\frac{\Delta v}{\Delta t}$. So the answer obtained from $\frac{\Delta v}{\Delta t}$ is the \emph{average acceleration} over the 20 seconds, which is not the same as the centripetal acceleration. As you made the $\Delta t$ smaller and smaller, it would converge to the value you got for $a_c$. So technically your original answer is correct (the $1.09$ one) if they asked for the magnitude of the centripetal acceleration. If they asked for the magnitude of the $\emph{average acceleration}$, then the latter one would be right.$

ohh okay I get it now, thanks man!

InteGrand · Mar 13, 2016

jathu123 · Mar 13, 2016

InteGrand said:
$\noindent Oh by the way, the average acceleration vector doesn't point towards the centre here. That happens to the \emph{centripetal acceleration} vector, which is the limit of the average acceleration vector as $\Delta t \to 0$. To the direction of the average acceleration vector approaches centripetal as $\Delta t\to 0$.$

oh, thanks for that info haha

small centripetal acceleration question (1 Viewer)

jathu123

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Green Yoda

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jathu123

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InteGrand

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jathu123

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InteGrand

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jathu123

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