# Homework Help: Average acceleration of the a particle in circular motion

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1. Oct 10, 2016

### decentfellow

1. The problem statement, all variables and given/known data
A particle of mass $m$ moves along a circle of radius $R$. The modulus of the average vector of the force acting on the particle over the distance equal to a quarter of the circle is:
1. zero if the particle moves with uniform speed $v$.
2. $\dfrac{\sqrt2mv^2}{\pi R}$ if the particle moves with the uniform speed $v$.
3. $\dfrac{2\sqrt2mv^2}{\pi R}$ if the particle moves with the uniform speed $v$.
4. $ma$, if the particle moves with constant tangential acceleration '$a$', the initial velocity being equal to zero.

2. Relevant equations
$$\vec{F}_{avg}=m\vec{a}_{avg}$$
$$\vec{a}_{avg}=\dfrac{\displaystyle\int{\dfrac{dv}{dt}}dt}{\displaystyle\int{dt}}=\dfrac{\displaystyle\int{dv}}{\displaystyle\int{dt}}=\dfrac{\displaystyle\int{a}dt}{\displaystyle\int{dt}}$$
$$\vec{a}_{avg}=\dfrac{\Delta v}{\Delta t}=\dfrac{v_f-v_i}{t_f-t_i}$$

Okay these are all the formulas that I know for the average acceleration (or that I have devised from the my knowledge), do write some more if you know.

3. The attempt at a solution
Let's consider the first case where the particle executes uniform circular motion, in this case the average acceleration will be $\vec{v}=\dfrac{\vec{v}_f-\vec{v}_i}{\Delta t}$

Consider the figure of the mentioned case given:-

As $\vec{a}_{avg}=\dfrac{\Delta v}{\Delta t}=\dfrac{-v\hat{j}+v\hat{i}}{\Delta t}$
Now, as the particle moves with same speed, so the time taken to cover quarter circle will be $\Delta t= \dfrac{\pi r}{2v}$

$$\therefore \vec{a}_{avg}=\dfrac{2v^2(\hat{i}-\hat{j})}{\pi r} \implies \left|\vec{a}_{avg}\right|=\dfrac{2\sqrt2v^2}{\pi r}$$

$\therefore \text{The average vector force acting on the particle over the distance equal to a quarter of the circle \\ if the particle executes uniform circular motion is} \implies |\vec{F}_{avg}|=\dfrac{2\sqrt2 mv^2}{\pi r}$

Now, consider the second case when the particle executes accelerated circular motion.

In this case there can be two approaches(that's what I think).

The first approach being that in which we consider $\vec{a}_{avg}=\dfrac{\Delta v}{\Delta t}$, which gives us the average acceleration magnitude as $|\vec{a}_{avg}|=\dfrac{at}{t}=a$.

Now, the second approach in which we consider the average acceleration as $|\vec{a}_{avg}|=\dfrac{\displaystyle\int{a}_{net}dt}{\displaystyle\int{dt}}$

Now, here the acceleration of the particle executing accelerated circular motion is given by $a_{net}=\sqrt{a_R^2+a_t^2}=\sqrt{\left(\dfrac{v^2}{r}\right)^2+a^2}=a\sqrt{\left(\dfrac{at^2}{r}\right)^2+1}$

Now I found out from wolframalpha that the integration $\int{a dt}$ id not doable by me, and the result is definitely something which doesn't result into average force vector to be $ma$.

So, please tell me where I am going wrong.

Edit:- Made a correction which is as follows:
$\vec{a}_{avg}=\dfrac{\displaystyle\int{a}dt}{\displaystyle\int{dt}} \rightarrow|\vec{a}_{avg}|=\dfrac{\displaystyle\int{a}_{net}dt}{\displaystyle\int{dt}}$

Last edited: Oct 10, 2016
2. Oct 10, 2016

### kuruman

$\vec{a}_{avg}=\dfrac{\displaystyle\int{a}dt}{\displaystyle\int{dt}}$
This is wrong. You show a vector equal to a scalar. The left side should the magnitude of the acceleration to be consistent with the right side. Considering that a is constant, the expression gives a = a, which you knew from the start.

3. Oct 10, 2016

### decentfellow

4. Oct 10, 2016

### kuruman

OK, but now you have to explain what you mean by $a_{net}$. It looks like you are averaging the magnitude of the instantaneous acceleration, so it really should be $a(t)$. For uniform circular motion, $a(t) = a = constant$. I don't see where you are going with this. It looks like you have already derived one of the possible answers. If you are wondering whether choice 4 could be correct, you need to consider if $ma$ could conceivably be "the modulus of the average vector of the force acting on the particle over the distance equal to a quarter of the circle" for a particle that has tangential acceleration $a$. You already have the pieces you need to answer this question.

5. Oct 10, 2016

### decentfellow

I am indeed averaging the instantaneous acceleration but I am also including the centripetal acceleration in doing so.

And I apologise for not mentioning that it is a more than one correct type question but surprisingly it has only one option correct which is option (c), but I think that (d) is also correct because of what I have shown in the first attempt in the accelerated circular motion part of my OP. And so I started thinking of ways to prove the option (d) as wrong just to see where I might be going wrong in my first attempt and hence I did the second attempt and now as you might have guessed I am totally confused.

6. Oct 10, 2016

### kuruman

You are understandably confused because the statement of question 4 is confusing. I am be tempted to say "deliberately confusing." This is why I think so. Here is question 4 as stated now

The modulus of the average vector of the force acting on the particle over the distance equal to a quarter of the circle is
$ma$ if the particle moves with constant tangential acceleration $a$.

The usual symbol for tangential acceleration is $a_t$ to distinguish it from the centripetal component $a_c$. Why is subscript "t" suppressed here? Suppose I rephrased the question to read,

The modulus of the average vector of the force acting on the particle over the distance equal to a quarter of the circle is $ma_t$ if the particle moves with constant tangential acceleration $a_t$.

It implies that the net force is mass times only one of its components (averaged or not) when actually there is another, centripetal, component. I suspect that the subscript was suppressed so that the unwary reader would see $ma$ and have the knee-jerk reaction of thinking "net force". That's what I meant by "deliberately confusing." The work you have shown and what you said about it convince me that your understanding of the physics behind this question is above the understanding that this question is designed to test.

7. Oct 10, 2016

### decentfellow

But then what about the first attempt that I made regarding this case, it (till now) seems perfectly fine to me, because it is also correct to define average acceleration as change in velocity over the time in which the change took place, in which case the first attempt seems fine to me.

Last edited: Oct 10, 2016
8. Oct 10, 2016

### kuruman

First attempt is fine and matches answer 3. That's what I meant earlier when I said,

9. Oct 10, 2016

### decentfellow

Not that one, I was referring to the first of the two attempts that I made regarding the accelerated circular motion of the particle.

10. Oct 10, 2016

### kuruman

You mean this one?
Again, you are equating a vector to a scalar.

11. Oct 10, 2016

### haruspex

@decentfellow , in getting to your first answer (option 3) you used the net change in the velocity (vector). Why are you not using that approach in assessing option 4?
The tricky part, perhaps, is finding the speed after a quarter turn, but it's not hard. You can think of the tangential movement as being one dimensional, with a known initial speed, constant acceleration, and known distance.

12. Oct 10, 2016

### decentfellow

Yeah I meant that one. And you are indeed correct it should be $\vec{a}_{avg}=\dfrac{\Delta \vec{v}}{\Delta t}$.

And as @haruspex pointed out, I should deal with them with the vector approach. Although, I dealt with them with that approach only, it was just that I did not show it very clearly here.

After completion of the quarter circle the direction of the velocity is in the -ve x-direction , and its magnitude can be found as follows:
As, the body has zero initial linear velocity(hence angular velocity also), so we have
$\dfrac{1}{2}\alpha t^2=\dfrac{\pi}{2} \implies t=\sqrt{\dfrac{\pi}{\alpha}}=\sqrt{\dfrac{\pi r}{a}}$

So, as the particle completes its quarter circle its linear velocity, is $\implies \omega=\alpha t \implies \omega=\dfrac{a}{r}\sqrt{\dfrac{\pi r}{a}}=\sqrt{\dfrac{\pi a}{r}} \\ \therefore v=r\omega=\sqrt{\pi r a}$

So, we get $\vec{v}=-\sqrt{\pi r a}\hat{i}$, therefore the average acceleration vector will be $\vec{a}_{avg}=\dfrac{\Delta \vec{v}}{\Delta t}=-\dfrac{\sqrt{\pi r a}}{\sqrt{\dfrac{\pi r}{a}}}\hat{i}=-a\hat{i}$. So, we get the magnitude of average acceleration as $|\vec{a}_{avg}|=a$, hence the magnitude of average vector of force is $|\vec{F}_{avg}|=m|\vec{a}_{avg}|=ma$.

This method does give the magnitude of average vector of force to be $ma$, but why does the method which included integration give such an other worldly answer (it certainly looked like that to me).

Also, I just noticed that I have made a blunder while indicating the velocity vectors in the diagram that I uploaded in the OP, so yeah just swap the direction of the vecotrs.

13. Oct 10, 2016

### haruspex

In that approach, you were integrating anet, taking that as the magnitude of the acceleration at time t. Integrating the magnitude of the acceleration will not give you the net acceleration (a vector). It's like confusing speed with velocity.

14. Oct 10, 2016

### decentfellow

So you are saying that what I did wring in my second attempt was that that I was integrating a scalar and equating it to a vector. I think I found out what you were implying so, based on that I think the average acceleration should be $\dfrac{\displaystyle\int{\left(-\dfrac{v^2}{r}\hat{r}+\vec{a}\right)}dt}{\displaystyle\int dt}$, am I right?

15. Oct 10, 2016

### haruspex

Yes, if $\vec a$ is defined as the tangential acceleration vector.

16. Oct 10, 2016

### decentfellow

Yes I by $\vec{a}$ I meant tangential acceleration and the seems like the result is same as that which I got in the first attempt.

What I did was $\vec{a}_{avg}=\dfrac{\displaystyle\int\left({-\left(\dfrac{a^2t^2}{r}\cos{\left(\dfrac{at^2}{2r}\right)}+a\sin{\left(\dfrac{at^2}{2r}\right)}\right)\hat{i}+\left(-\dfrac{a^2t^2}{r}\sin{\left(\dfrac{at^2}{2r}\right)}+a\cos{\left(\dfrac{at^2}{2r}\right)}\right)\hat{j}}\right)dt}{\displaystyle \int dt} \\ =a(\cos{\dfrac{at^2}{2r}}\hat{j}-\sin{\dfrac{at^2}{2r}}\hat{i})$

And the magnitude of average acceleration is $a$.

Now I am pretty sure that the 4th option is definitely correct.

17. Oct 11, 2016

### ehild

It is right. As you said previously,
$\vec{a}_{avg}=\dfrac{\vec \Delta v}{\Delta t}=\dfrac{\vec v_f - \vec v_i}{t_f-t_i}$
As $\vec v_i=0$, $\vec{a}_{avg}=\dfrac{\vec v_f}{t_f-t_i}$.
The magnitude of the average acceleration is the final speed divided by the time, and the speed is vf=a(tf-ti), as the tangential component of acceleration is constant.