I really appreciate the help.Finding Force in Nonuniform Circular Motion

[phyzz]

Homework Statement

Let a body with a mass m move along a circle of radius R. The absolute value (magnitude) of its velocity changes linearly in time from v to 2v during one complete rotation. Find the absolute value of the force that act on this body as a function of time.

The Attempt at a Solution

I first found the tangential acceleration:

<br /> a_{tan} = {\frac{∆v}{∆t}}<br />
<br /> a_{tan} = {\frac{2v - v}{t}} = {\frac{v}{t}}<br />

And for the radial acceleration I have
<br /> a_{R} = {\frac{v^2}{R}}<br />

Overall acceleration
<br /> \|\vec{a}\| = \sqrt{a_{tan}{}^{2} + a_{R}{}^{2}} = \sqrt{{\frac{v^2}{t^2}} + {\frac{v^4}{R^2}}}<br />

So by using Newtons 2nd law
<br /> ∑\vec{F}\ = m\vec{a}\<br />

I get
<br /> |F| = m\sqrt{{\frac{v^2}{t^2}} + {\frac{v^4}{R^2}}}<br />

I don't feel confident at all with my response, the question does ask as a function of time and I have v's, m and an R in there. Have I gone completely wrong somewhere?

Thank you

 

[voko]

The general line of thinking is correct. Unfortunately, you did not compute the tangential acceleration correctly. Note the description said the velocity changed in the specified way during one revolution. "One revolution" is not reflected in your equation for tangential acceleration in any way; besides, the equation does not seem to make much sense, as the acceleration, according to it, gets smaller as time goes on, which is definitely not prescribed by the description.

 

[phyzz]

The general line of thinking is correct. Unfortunately, you did not compute the tangential acceleration correctly. Note the description said the velocity changed in the specified way during one revolution. "One revolution" is not reflected in your equation for tangential acceleration in any way; besides, the equation does not seem to make much sense, as the acceleration, according to it, gets smaller as time goes on, which is definitely not prescribed by the description.

Ok now I have
<br /> a_{tan} = {\frac{dv}{dT}}<br />
so I differentiate
<br /> v = {\frac{2πR}{T}}<br />
where T is the period, and I get
<br /> v&#039; = a_{tan} = 2πRln(T)<br />
And for the radial acceleration I have
<br /> a_{R} = {\frac{v^2}{R}} = {\frac{4π^2R}{T^2}}<br />

Overall acceleration
<br /> \|\vec{a}\| = \sqrt{a_{tan}{}^{2} + a_{R}{}^{2}} = \sqrt{4π^2R^2(ln(T))^2 + {\frac{16π^2R^2}{T^2}}} = {\frac{2πR}{T}}\sqrt{T^2(ln(T))^2 + 4}<br />

So by using Newtons 2nd law
<br /> ∑\vec{F}\ = m\vec{a}\<br />

I get
<br /> |F| = \frac{2πRm}{T}\sqrt{T^2(ln(T))^2 + 4}<br />

Is this better or worse? Thanks for your help

 

[voko]

<br /> v = {\frac{2πR}{T}}<br />
where T is the period

This is true only for circular constant-speed motion. Which is not the case here.

We do, however, know that the speed increases linearly in time. What does that mean about tangential acceleration? Hint: consider straight-line motion with similar properties.

 

[phyzz]

This is true only for circular constant-speed motion. Which is not the case here.

We do, however, know that the speed increases linearly in time. What does that mean about tangential acceleration? Hint: consider straight-line motion with similar properties.

Oh, I see. In that case I have:

<br /> a_{tan} = {\frac{dv}{dT}}<br />
I have:
s = 2πR (circumference)
u = v (initial velocity)
v = 2v (final velocity)
a = ? (what we're trying to find)
t = / (not used in this case)
<br /> v^2 = u^2 + 2as<br />
<br /> 4v^2 = v^2 + 4πRa_{tan}<br />
<br /> 3v^2 = 4πRa_{tan}<br />
and I finally get
<br /> a_{tan} = {\frac{3v^2}{4πR}}<br />

And for the radial acceleration I have (subbing in the period equation with T)
<br /> a_{R} = {\frac{v^2}{R}} = {\frac{4π^2R}{T^2}}<br />

Overall acceleration
<br /> \|\vec{a}\| = \sqrt{a_{tan}{}^{2} + a_{R}{}^{2}} = \sqrt{\frac{9v^4}{16π^2R^2} + {\frac{16π^4R^2}{T^4}}} = {\frac{3}{4T^2}}\sqrt{\frac{9v^4T^4 + 16π^4R^2}{9π^2R^2}}<br />

So by using Newtons 2nd law
<br /> ∑\vec{F}\ = m\vec{a}\<br />

I get
<br /> |F| = {\frac{1}{4πRT^2}}\sqrt{9v^4T^4 + 16π^4R^2}<br />

 

[voko]

Again, there is no T because the motion is not periodic.

With the tangential acceleration known, you can express the speed as a function of time, so you should be able to express the radial acceleration as a function of time.

 

[phyzz]

Again, there is no T because the motion is not periodic.

With the tangential acceleration known, you can express the speed as a function of time, so you should be able to express the radial acceleration as a function of time.

So I integrate my atan which is dv/dt by separating the variables, find v which gives me
v = -4πR/3t

and then I sub this into the aR equation:

aR = v^2/R
effectively getting rid of the - sign by squaring the v and I get
aR = 16π^2R^2/9t^2

and then I find the norm to get overall acceleration and sub that into F = m||a||?

 

[voko]

Your tangential acceleration is constant; you got it by ASSUMING it was constant, otherwise the formulae you used to derive it would have been incorrect. The "v" you have there is not the the time varying speed; it the initial speed "v" as per the description of the problem. You should rethink your notation to avoid this confusion.

 

[phyzz]

Your tangential acceleration is constant; you got it by ASSUMING it was constant, otherwise the formulae you used to derive it would have been incorrect. The "v" you have there is not the the time varying speed; it the initial speed "v" as per the description of the problem. You should rethink your notation to avoid this confusion.

I no longer know how to approach this problem

 

[voko]

Consider straight-line motion again: if the initial velocity is v, and the constant acceleration is a, what is the velocity at time t? Is that different from circular motion with constant tangential acceleration?

 

[phyzz]

Consider straight-line motion again: if the initial velocity is v, and the constant acceleration is a, what is the velocity at time t? Is that different from circular motion with constant tangential acceleration?

Is tangential accn constant?

 

[voko]

Is tangential accn constant?

If it was not, then how could you use SUVAT to find it? What in the problem description indicates it is constant?

 

[phyzz]

If it was not, then how could you use SUVAT to find it? What in the problem description indicates it is constant?

So it is constant?

 

[voko]

So it is constant?

Yes, I have said this multiple times. But do you understand why? You should go back to #4 and think about it.

 

[phyzz]

Yes, I have said this multiple times. But do you understand why? You should go back to #4 and think about it.

EDIT: Is it constant because the speed changes 'linearly' as per the q?

Ok, thank you. I have now integrated my atan and got vtan = -4πR/3t

I know atan is (vtan)'

so (vtan)' = 4πR/3t^2

and (vtan)'^2 = 16π^2R^2/9t^4

add that on to aR^2 = 16π^2R^2/T^2

and sqrt all of that to get ||a||?

then use F = m||a||

 

[voko]

EDIT: Is it constant because the speed changes 'linearly' as per the q?

Ok, thank you. I have now integrated my atan and got vtan = -4πR/3t

Do you understand what "changes linearly in time" means? What is a linear function?

 

[phyzz]

Do you understand what "changes linearly in time" means? What is a linear function?

I'm not sure

 

[voko]

I see now.

A linear function is one whose graph is a straight line. What is the general form of such a function?

 

[phyzz]

I see now.

The answer?

A linear function is one whose graph is a straight line. What is the general form of such a function?

y = mx + c

 

[voko]

Now that you know what a linear function is, what is the time derivative of a linear function of time?

 

[phyzz]

Now that you know what a linear function is, what is the time derivative of a linear function of time?

Velocity?

 

[voko]

What is the time derivative of ANY linear function of time, not necessarily position?

 

[phyzz]

What is the time derivative of ANY linear function of time, not necessarily position?

I don't know

 

[voko]

Are you saying you cannot differentiate what you wrote in #19?

 

[phyzz]

Are you saying you cannot differentiate what you wrote in #19?

y' = m ?

 

[voko]

So what is the time derivative of a linear function of time? How does it depend on time?

 

[phyzz]

So what is the time derivative of a linear function of time? How does it depend on time?

I really don't know. Is the integration and then differentiation approach wrong?

 

[voko]

I really don't know. Is the integration and then differentiation approach wrong?

I am not sure what you mean by that. To solve this problem, you have to understand what a linear function is. It is a very basic notion, predating calculus. You have to understand what its derivative is. You have in fact derived that the derivative is f'(t) = m, but you are unable to say what this means. This really baffles me. Is m a constant, a variable, another function of time?

What does this all mean if the linear function f(t) is velocity and f'(t) then is acceleration? And if you know acceleration, what is then the velocity?

 

[phyzz]

I am not sure what you mean by that. To solve this problem, you have to understand what a linear function is. It is a very basic notion, predating calculus. You have to understand what its derivative is. You have in fact derived that the derivative is f'(t) = m, but you are unable to say what this means. This really baffles me. Is m a constant, a variable, another function of time?

What does this all mean if the linear function f(t) is velocity and f'(t) then is acceleration? And if you know acceleration, what is then the velocity?

Could you please start from the beginning because I am now more confused than before starting the problem

 

[voko]

We are back at the very beginning. The problem says that the magnitude of velocity is a linear function of time. Let's denote this f(t) - not v because v is used as a constant specification of initial velocity, so f(0) = v. Tangential acceleration is the time derivative of velocity, i.e., f'(t). What is it?

 

[phyzz]

We are back at the very beginning. The problem says that the magnitude of velocity is a linear function of time. Let's denote this f(t) - not v because v is used as a constant specification of initial velocity, so f(0) = v. Tangential acceleration is the time derivative of velocity, i.e., f'(t). What is it?

s = /
u = v
v = 2v
a = ?
t = t

v = u + at
v = at
a = v/t

f'(t) = v/ t
1/v dv = 1/t dt
ln(v) = ln(t) + C
t = 0 when v = 0
I get infinite C?

 

[voko]

In #19 and #25 you wrote what f(t) and f'(t) are. You don't need to solve any differential equations. You just need to determine the constants involved, that's all there is to it.

 

[phyzz]

In #19 and #25 you wrote what f(t) and f'(t) are. You don't need to solve any differential equations. You just need to determine the constants involved, that's all there is to it.

I don't know what you mean

What about the period of one rotation? You said it had to be included somewhere

 

[voko]

There is no period. The motion is not uniform.

I do not think this is going in the right direction. I do not feel that you understand the significance of linearity of velocity. I will make one final attempt.

The magnitude of the velocity is a linear function of time. That means $v(t) = a_{\tau}t + v_0$. The derivative is $v'(t) = a_{\tau}$. $a_{\tau}$ is tangential acceleration, $v_0$ is the initial velocity, which is denoted just as $v$ in the problem, which is very unfortunate because it got you badly confused earlier. You have already deduced $a_{\tau}$ in terms of $R$ and $v_0$ (again, it was confusingly denoted simply as $v$). With that, you can easily restore $v(t)$ and thus obtain $a_n$, the normal (or radial) acceleration. Then it is trivial to get the magnitude of net acceleration and force.

 

[phyzz]

There is no period. The motion is not uniform.

I do not think this is going in the right direction. I do not feel that you understand the significance of linearity of velocity. I will make one final attempt.

The magnitude of the velocity is a linear function of time. That means $v(t) = a_{\tau}t + v_0$. The derivative is $v'(t) = a_{\tau}$. $a_{\tau}$ is tangential acceleration, $v_0$ is the initial velocity, which is denoted just as $v$ in the problem, which is very unfortunate because it got you badly confused earlier. You have already deduced $a_{\tau}$ in terms of $R$ and $v_0$ (again, it was confusingly denoted simply as $v$). With that, you can easily restore $v(t)$ and thus obtain $a_n$, the normal (or radial) acceleration. Then it is trivial to get the magnitude of net acceleration and force.

Thank you for your patience. So the radial accn would be aR = v(t)^2 / R

v(t) being what was found earlier from $v(t) = a_{\tan}t + v_0$