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enter image description here What is the difference between centripetal acceleration and average acceleration in this worked example?
If $\frac{\Delta v}{\Delta t}=\frac{v^2}{r}$ and $v=3$ m/s change in time is $1$ second and $r$ is $1$ m why change in tangential velocity is not $9$ m/s$^2$ but $6$ m/s?

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    $\begingroup$ Pedagogically, asking about average acceleration in this situation (uniform circular motion) is nonsense! Why do books do things like this? There are better examples of finding average acceleration, e.g., top fuel dragster timings. $\endgroup$ – Bill N Mar 19 '18 at 15:22
  • $\begingroup$ Why do you think that $\Delta v / \Delta t = v^2/r$? That's not true. $\endgroup$ – Bill N Mar 19 '18 at 15:24
  • $\begingroup$ It is written in my book $\endgroup$ – Marva Jami Mar 19 '18 at 15:51
  • $\begingroup$ Average accn .mean change in velocity /change in time $\endgroup$ – Yuvraj Singh... Sep 10 '19 at 13:14
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The centripetal acceleration is the instantaneous acceleration which is given by $$\vec{a}_{\rm ins}=-\frac{v^2}{r}\hat{r}$$ which has a magnitude of $9$ in your case, but is constantly changing in direction.

The average acceleration is defined by $$\vec{a}_{\rm ave}=\frac{\Delta \vec{v}}{\Delta t}$$ which has magnitude of $6$ in your case and is pointing to the right.

By definition $$\vec{a}_{\rm ins}(t)=\lim_{\Delta t \to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\lim_{\Delta t \to 0}\vec{a}_{\rm ave}$$ or equivalently $$\vec{a}_{\rm ave}=\frac{1}{\Delta t}\int_t^{t+\Delta t}\vec{a}_{\rm ins}dt$$

In your case, you can interpret it this way: Although the instantaneous centripetal acceleration has a constant magnitude of $9$, it is constantly changing in direction. Hence there are some cancellations when you find the vector sum in the integral above, which leads to an average acceleration of magnitude $6$ pointing to the right.

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  • $\begingroup$ I haven't studied calculus can you explain in it in a way which doesn't involve calculus by the way it is written in my book that Δv/Δt=v^2/r in this problem if i plug the values of v,r,and Δt the change in velocity will be 9m/s but in the book it is 6m/s $\endgroup$ – Marva Jami Mar 19 '18 at 12:36
  • $\begingroup$ So to find the average acceleration you just need to subtract the initial velocity vector from the final velocity vector, then divide it by time taken, and you get 6 to the right. That's the definition. $\endgroup$ – velut luna Mar 19 '18 at 12:37
  • $\begingroup$ If you haven't studied calculus, you can't understand instantaneous centripetal acceleration. But you are asking about it. I don't know how to explain without calculus. Sorry! $\endgroup$ – velut luna Mar 19 '18 at 12:38
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Here another way to solve this problem.

As explained in some of the previous answers, the centripetal acceleration is a vector pointing to the center of the circle or normal to the velocity vector. As the toy car is moving down the circle, counterclockwise, from top position (π/2) to the bottom position (-π/2), the acceleration vector is moving counterclockwise as well, from -π/2 to π/2, as shown below:

enter image description here

The average acceleration would be a vector some of all these vectors divided by the total angle π/2 –(-π/2)= π.

As can be seen from the picture, the vertical components of these vectors (blue and red) will cancel each other, while horizontal components (green, pointing to the right) will add up. Therefore, the sum of all vectors will be pointing to the right.

The horizontal component of a vector with an absolute value 9 could be calculated as 9*cos(x), where the x is angle, therefore the average acceleration magnitude could be calculated as follows:

enter image description here

This is not exactly 6, but close:)

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Average (or mean) acceleration (over a period of time) is defined by$$\text{mean acceleration}=\frac{\text{final velocity-initial velocity}}{\text{time taken to change}}$$You'll see that this is exactly what has been used in the answers to (b) and (c ). Note that the subtraction is a vector subtraction, as shown in the diagram.

The acceleration or instantaneous acceleration at a particular time is the limiting value of the mean acceleration as the time interval tends to zero. We choose a time interval centred on the time at which we want the acceleration. As the bottom line of the fraction above gets smaller, so does the top line, but the value of the fraction itself tends not to zero, but to this so-called limiting value.

To understand what the last paragraph means, it's well worth doing some calculations. Consider time intervals of, perhaps, 0.50 s, 0.25 s, 0.125 s and calculate the mean acceleration for each. You'll need different vector diagrams each time, in order to to the subtractions. The magnitude of the mean acceleration will get closer and closer to $\frac{v^2}{r}$ [If you're going to do this, you'll need to use a value for the radius correct to at least 3 sig figs, that is $\frac{3.00\ \text {m s}^{-1}\times 2.00 \text{s}}{2 \pi}=0.955$ m, and indeed to do all your working to 3 or 4 sig figs.]

We can show formally that the limiting value has a magnitude of $\frac{v^2}{r}$ and is directed towards the centre of the circle. The derivation is to be found in textbooks.

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