# Circular motion/conical pendulum?

1. Dec 11, 2011

### jiboom

one end of a light inelastic string is attached to a point A vertically above a point O on a smooth horizontal plane and at a height h above it. The string carries a particle P of mass m at its other end. WHEN JUST TAUT the string is inclined to the vertical at an angle S

if P moves in a horizontal circle,centre O,with speed v,show that

v^2<= hg tan^2 s

im reading this as the particle is on the smooth plane,as the circle centre is O,so my equations of motion are,with T=tension,r=rad of circle,R= reaction force

R+Tcos S-mg=0
Tsin S=mv^2/r

i know r=htan S

i know at speed V string is just taut,but putting T=0 makes no sense, so how to proceed?if i put T>0 i still need to get rid of the reaction force.

regarding the just taut part. on another question i had 2 strings attached to a particle at one end, and the other ends of the strings were attached at points in a vertical line. the paricle moved in circle at a speed which just made strings taut. to do the question i had to let the tension in the bottom string = 0. so what does just taut mean?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 11, 2011

### PeterO

Imagine you make an amusement ride, with a small/toy aeroplane tethered to the top of a pole, but merely travelling around on the ground.
You could have the aeroplane travel slowly in a circle, keeping the tether tight.
The centripetal force required would be small, as the aeroplane is traveling slowly. If that centripetal force is supplied by the tether, then the tether would be also supplying a small vertical force on the aeroplane , so that the reaction force from the ground would be smaller than if the aeroplane was stationary.
If the aeroplane travels faster, the force supplied by the tether has to increase, to provide a larger centripetal force, and so a larger vertical force as well, so the reaction force would reduce.
There will exist a speed where the reaction force would be reduced to zero.
Travel any faster and the aeroplane would have to take off [and fly in a circle of larger radius, but less distance below the top of the pole]

3. Dec 11, 2011

### PeterO

For ease of explanation I will assign some measurements - so I can refer to them.

I view this question as , for example, a 2m string attached to the top of the pole, and a 1.5 m string attached to a point 1m lower. A mass is attached to the other end of each string

You can draw a profile diagram using a compass and ruler

The 2m string means the mass can only be in a position 2 m [or less] from the top of the pole - so draw a 3m vertical line, then a quarter circle arc, radius 2, centred at the top of that line.

The 1.5m string means the mass can only be 1.5m, or less from a point 1m below the top of the pole - so add a second quarter circle, radius 1.5m, centred at 1m below the top of the pole.

[You may want to do your diagram in cm to fit it onto a sheet of paper]

You can now identify possible positions of the mass: at first somewhere on the 2m arc and then, from the point of intersection, somewhere on the 1.5m arc.

If the particle is hanging , stationary, the 2m string will be taught, and the 1.5m string will be limp [as its ends are only 1 m apart]. The mass will be on the low point of the 2m arc in your diagram.

If the particle moves slowly in a circle, it may be making a circle of radius 0.5m.
The 2m string will be tight, and the 1.5m string will still be limp. the mass will be at a point on the 2m arc.

As the speed increases, eventually both strings will be taught: the mass will be at the point of intersection of the two arcs.

You can use geometry [cosine rule] to show that at that point, the 2m string makes an angle of 46.5675o with the pole.

At first, the tension in the 2m string will be large, while the 1.5m string will have a tension of approx zero. You could calculate the speed necessary for this situation to exist.

If the speed is increased slightly, the tension in the 1.5m string will increase and the tension in the 2m string will drop.
Together the tensions supply just enogh force to keep tha mass circling at the point.

Increase further and we reach a point where the tension in the 2m string has dropped to zero, and the mass is entirely supported in its motion by the 1.5 m string.

Go a little faster and the mass will move to a position on the 1.5m arc, where the 1.5 m string is tight and the 2m string is slack.

If the mass travels really fast, the 1.5m string will be almost horizontal. TA that poiint the mass is only 1.8m from the top of the pole [use pythagorus] showing the 2m string is certainly slack.

In your problem you should have set the tension in each string to zero in turn and could get two different speeds as an answer. The lower speed means bottom string almost slack, the higher speed means top string almost slack.

4. Dec 12, 2011

### jiboom

thanks for the replies.

so if im getting this right, here the just taut means R=0?
then ]

Tcos S-mg=0
Tsin S=mv^2/(h tan S)

but i still cant get the condition.

i can divide the 2 to get

tan S=v^2/(hg tan S)

giving v^2=hg tan^2 S

how do i get the less than part?

5. Dec 13, 2011

### jiboom

is this ok reasonong?

R+Tcos S-mg=0
Tsin S=mv^2/r

R=mg-Tcos S

we have R>=0

T<=mg/cos S

and
T=mV^2/rsin S

so

mV^2/rsin S<=mgcos S

v^2<ghtan^2 S

6. Dec 13, 2011

### PeterO

Not sure where tan2S comes from in your substitutions, but substitutions along those lines give the expression.

Don't forget than r = h.tanS from the geometry of the situation.