# Homework Help: Nonuniform Circular Motion

1. Oct 15, 2012

### phyzz

1. The problem statement, all variables and given/known data

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.

3. The attempt at a solution

I first found the tangential acceleration:

$$a_{tan} = {\frac{∆v}{∆t}}$$
$$a_{tan} = {\frac{2v - v}{t}} = {\frac{v}{t}}$$

And for the radial acceleration I have
$$a_{R} = {\frac{v^2}{R}}$$

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

So by using Newtons 2nd law
$$∑\vec{F}\ = m\vec{a}\$$

I get
$$|F| = m\sqrt{{\frac{v^2}{t^2}} + {\frac{v^4}{R^2}}}$$

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

2. Oct 15, 2012

### 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.

3. Oct 15, 2012

### phyzz

Ok now I have
$$a_{tan} = {\frac{dv}{dT}}$$
so I differentiate
$$v = {\frac{2πR}{T}}$$
where T is the period, and I get
$$v' = a_{tan} = 2πRln(T)$$
And for the radial acceleration I have
$$a_{R} = {\frac{v^2}{R}} = {\frac{4π^2R}{T^2}}$$

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

So by using Newtons 2nd law
$$∑\vec{F}\ = m\vec{a}\$$

I get
$$|F| = \frac{2πRm}{T}\sqrt{T^2(ln(T))^2 + 4}$$

Is this better or worse? Thanks for your help

4. Oct 15, 2012

### voko

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.

5. Oct 16, 2012

### phyzz

Oh, I see. In that case I have:

$$a_{tan} = {\frac{dv}{dT}}$$
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)
$$v^2 = u^2 + 2as$$
$$4v^2 = v^2 + 4πRa_{tan}$$
$$3v^2 = 4πRa_{tan}$$
and I finally get
$$a_{tan} = {\frac{3v^2}{4πR}}$$

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

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

So by using Newtons 2nd law
$$∑\vec{F}\ = m\vec{a}\$$

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

Last edited: Oct 16, 2012
6. Oct 16, 2012

### 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.

7. Oct 16, 2012

### phyzz

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||?

8. Oct 16, 2012

### 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.

9. Oct 16, 2012

### phyzz

I no longer know how to approach this problem

10. Oct 16, 2012

### 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?

11. Oct 16, 2012

### phyzz

Is tangential accn constant?

12. Oct 16, 2012

### voko

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

13. Oct 16, 2012

### phyzz

So it is constant?

14. Oct 16, 2012

### voko

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

15. Oct 16, 2012

### phyzz

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||

16. Oct 16, 2012

### voko

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

17. Oct 16, 2012

### phyzz

I'm not sure

18. Oct 16, 2012

### voko

I see now.

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

19. Oct 16, 2012

### phyzz

y = mx + c

20. Oct 16, 2012

### voko

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

21. Oct 16, 2012

### phyzz

Velocity?

22. Oct 16, 2012

### voko

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

23. Oct 16, 2012

### phyzz

I don't know

24. Oct 16, 2012

### voko

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

25. Oct 16, 2012

y' = m ?