Homework Help: Motion of Vertical Circle

1. Jul 24, 2016

Taniaz

Motion in a vertical circle*
1. The problem statement, all variables and given/known data

Case 1: When a particle P of mass m moves in a vertical circle centered on O, its motion is governed by two forces, its weight mg and a force R directed radially inwards constraining it to move in a circle radius r. Resolving parallel to an perpendicular to OP (radially inwards and tangentially respectively) where OP is at an angle theta to the downward vertical, and applying Newton's Second Law, we get:

R-mg cos(theta) = mv^2/r and mg sin(theta) = -m dv/dt

Case 2:
If the particle is constrained to circular motion due to R radially outwards, the forces are directed differently as follows (theta is measured with respect to the upwards vertical)

mg cos (theta) - R = mv^2/r and mg sin (theta) = m dv/dt

2.Relevant equations
As above

3.The attempt at a solution
After drawing the relevant diagrams, I was able to derive the 1 equation of both cases but I was a little unsure on how they got the second equation in both cases. I know they've use Newton's second law in this case but I don't know how they got the negative sign in the first case and a positive sign in the second case. Also, why did the angle change when the radial force changed direction? How can a force be radially outward?

2. Jul 24, 2016

PeroK

Circular motion requires a radially inwards force. A particle cannot move in a circle with a net outward force.

3. Jul 24, 2016

Taniaz

Yes I understand so the net force must be inwards.

Do you know how they derived mg sin (theta) = -m (dv/dt) for the R inwards case and mg sin (theta) = m dv/dt for the R outwards case?

4. Jul 24, 2016

CWatters

I think the only difference is which direction they define as "positive".

If you define "up" as positive then centripetal force acting towards the centre is negative.
If you define "down" as positive then centripetal force acting towards the centre is positive.

5. Jul 24, 2016

PeroK

The description is a bit of a mess and needs a diagram. I worked out the direction of the forces and then worked out what the text meant in terms of what was positive or negative.

And, the direction of the axes and the definition of theta changes between the two cases.

In the first case theta is 0 at the bottom and increases anticlockwise. In the second case theta is 0 at the top.

6. Jul 24, 2016

Taniaz

I wrote it exactly like it was written in the book.

Why is this the case "In the first case theta is 0 at the bottom and increases anticlockwise. In the second case theta is 0 at the top."?

7. Jul 24, 2016

PeroK

8. Jul 24, 2016

Taniaz

Well it has something to do with R being inwards and outwards?

9. Jul 24, 2016

Taniaz

This is the diagram I have for both cases. How is the horizontal component mg sin(theta) = -m dv/dt?
And in the second case, how is the horizontal component mg sin(theta) = m dv/dt?
We don't even know which direction v is in? It's not specified whether it's going upwards or coming downwards??

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10. Jul 24, 2016

Taniaz

If for R being radially inwards they mean that, for example, you have an object tied to the bottom of a string and you leave it hanging as it is then start to move it at an angle from the bottom upwards then it's actually working against gravity so it's going to oppose the horizontal component of weight which is trying to bring it back downwards. In case two you reverse the orientation so that the object is now at the top and you let it go from there. That way it is not opposing gravity and actually in the direction of the horizontal component of weight? I don't know.

11. Jul 24, 2016

Taniaz

Anyone?

12. Jul 24, 2016

Taniaz

I think if we use Newton's second law in the case of R being radially inwards, Σ Fy = mv^2/r and ΣFx= ma
In the case of R being radially inwards, work is being done against gravity so the only x-component present is mg sin (theta) so it will equal -m dv/dt
and vice versa for R being radially outwards.

Can someone just explain to me HOW R can be radially outwards in vertical motion??

13. Jul 24, 2016

IanBerkman

When it is constrained to the motion of a circle, R should not be a constant force radially inward/outward since the $mg$ term already has a component towards the bottom. For instance in the radially inward case, when the particle is at the bottom of the circle it needs to have a larger radially inward force than when it is on top to prevent it from 'leaving' the circle.
You get the second equation by looking at the velocity in the tangential direction (actually this is the only direction of the velocity). In the first case the velocity is in anti-clockwise direction and the second case in the clockwise. The $m\frac{dv}{dt}$ is then the force associated with this velocity.

14. Jul 24, 2016

Taniaz

Thank you for your reply. So when it is at the top, it needs a larger radially outward force or is it just one case where R is radially outwards at the top? Can it be radially outwards closer to the bottom? Does this also mean that whatever the radial component is, the net force's direction should always be inwards?

Is my reasoning for the second equation correct then? Sum of any tangential components= ma where dv/dt is a. In the radially inward case, -m dv/dt because work is being done against gravity?

Last edited: Jul 24, 2016
15. Jul 24, 2016

Taniaz

In addition to my previous post, please take a look at this example. I don't understand how they got an answer for both radially inward and outward .

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16. Jul 24, 2016

CWatters

I think the original statement is badly worded. To move in a circle there must be a net centripetal force acting inwards but there can be forces acting in lots of different directions. Consider an aircraft performing consecutive outside loops. We usually consider that there are four forces acting on the aircraft... Lift, Drag, Thrust and Weight. In order to move in a circle they must have a centripetal component = mv2/r.

Have a think about the Lift force at the very top of the outside loop. At that point the lift force can still be upwards (outwards) but it must be less than the weight if the aircraft is going to descend into another loop.

17. Jul 24, 2016

Taniaz

Thank you for your reply. That made is very clear but I just have one concern, can you take a look at the example picture I posted right before your post? How did they decide the reaction force of the wire on the bead was going to be radially inward? And then they later got both forces for inward and outward?

18. Jul 24, 2016

CWatters

PS: A lot depends on the speed of the aircraft. If the aircraft is very slow then at the top the lift force can be outwards (otherwise gravity would provide too much centripetal force). If the aircraft is a very fast jet then it might need to be inwards (gravity alone being unable to provide the required centripetal force).

19. Jul 24, 2016

Taniaz

I think in the example I posted, the action of the wire on the bead would have been radially outwards as per the book on the two cases and the reaction would have been radially inward? And because action and reaction are equal in magnitude but opposite in direction, magnitude of R outward = R inward. That's all I can think of.

20. Jul 24, 2016

CWatters

Will get back to you.

21. Jul 24, 2016

CWatters

Ok it's not too hard..

They started by making the assumption that the reaction force R is a positive force acting towards the centre.

Then when they put theta = 45 degrees into the equation...

R = 0.02g(2-3cos(theta))

..they get a negative answer so R must actually act outwards.

Did you follow how they got the equations ok?

22. Jul 24, 2016

CWatters

As for the force at 135 degrees... They use the CAST rule to find that Cos(135) is -ve. That makes R a positive value = inwards.

23. Jul 24, 2016

Taniaz

Yes I worked through the equations and understood them. Just reading your explanation now.

24. Jul 24, 2016

Taniaz

Ok that makes sense! Had they found that if, hypothetically speaking, cos 135 was positive which gave an answer of negative for R in the inwards direction then we would say that no it would be outwards as it is positive in the opposite direction? (hypothetically speaking)

25. Jul 24, 2016

CWatters

Yes. They start by defining positive R = inwards, so any time that equation gives a positive (negative) result the force is inwards (outwards).

As an exercise you could try assuming positive R = outwards. The equation will be slightly different but you should get the same answer when you substitute the values for theta in the end.

You can also easily work out the angle at which the force on the wire is zero (I make it about 48 degrees).