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decentfellow
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Homework Statement
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:
- zero if the particle moves with uniform speed ##v##.
- ##\dfrac{\sqrt2mv^2}{\pi R}## if the particle moves with the uniform speed ##v##.
- ##\dfrac{2\sqrt2mv^2}{\pi R}## if the particle moves with the uniform speed ##v##.
- ##ma##, if the particle moves with constant tangential acceleration '##a##', the initial velocity being equal to zero.
Homework 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.
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}}##
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