Force of gravity along the radial direction

[Pochen Liu]

I am trying to work out the velocity of the ball in a loop in terms of theta from the horizontal (like a unit circle) as it loses contact with the track. And having a lot of trouble understanding this equation

m*g*sin(theta) = m*v²/r

and this explanation:

The ball will leave the rail when the rail reaction is zero. This happens when the centripetal force needed to travel around the loop is totally supplied by the component of the gravity force directed towards the centre of the motion.

If I were trying to find the minimal velocity so that it will remain in contact throughout the whole loop, yes I understand that Fg = Fc and they act in the same direction, but if the ball is not directly upside down such as 45 degrees I do not understand how there is any Fg towards the centre of the circle because gravity only acts downwards

I tried working this out and by drawing a vector diagram I get

m*g = (m*v²/r) * sin(theta)

Because I am making the vertical component of Fn equal to Fg, as that is when it will start to fall. But this is wrong obviously, and I'm realising how bad my intuition is.

 

[jbriggs444]

Because I am making the vertical component of Fn equal to Fg

It falls away from the sides when the normal force is zero, not when the [downward component of] the normal force happens to be equal to the downward force of gravity.

The normal force is zero when the radial acceleration of the ball is equal to its radial acceleration due to gravity alone. Since gravity pulls at an angle that is not purely radial, the sin(theta) factor is needed.

A cos(theta) factor could describe the remaining tangential acceleration due to gravity, but we are not concerned with that -- it is at right angles to the normal force.

 

[256bits]

Hi Pochen,

I do not understand how there is any Fg towards the centre of the circle because gravity only acts downwards

A conundrum then, isn't it?
Have you dealt with boxes sliding down a ramp.
The weight of the box due to the gravitational force acts downwards vertically.
Yet, the normal force can be found.
This situation is similar.

 

[Pochen Liu]

The normal force is zero when the radial acceleration of the ball is equal to its radial acceleration due to gravity alone. Since gravity pulls at an angle that is not purely radial, the sin(theta) factor is needed.
.

I see where I went wrong, but I'm struggling to visualise what you mean by 'Since gravity pulls at an angle that is not purely radial, the sin(theta) factor is needed.'
Could you possibly give me a visual indication of what you mean?

 

[Pochen Liu]

Hi Pochen,

A conundrum then, isn't it?
Have you dealt with boxes sliding down a ramp.
The weight of the box due to the gravitational force acts downwards vertically.
Yet, the normal force can be found.
This situation is similar.

I have done banked curve questions, however I can't get my vector diagrams to work with this simple question for some reason. Because this radial thing is confusing me, with the banked curve I know the vertical component of Fn must be Fg etc

 

[256bits]

I have done banked curve questions, however I can't get my vector diagrams to work with this simple question for some reason. Because this radial thing is confusing me, with the banked curve I know the vertical component of Fn must be Fg etc

Perhaps if you start with the ball at the lowermost position and resolve the vector components of gravity along the radial and tangential components as the ball moves along the track, to get an idea of what is going on.

At the lowermost position, call that position 0 degrees for sake of argument all of the weight will be directed downwards onto the track.
As the ball goes around the track:
Between lowermost and before 90 degrees some of the weight will be directed radially onto the track, and some tangentially.
At 90 degrees, all weight will be tangential.
After 90 degrees and onwards, ( up to 270 degrees on the other side passing by the topmost position), the weight will have a radial component way from the track.
The v^2 r has to compensate, or the ball falls away from the track.

Just relate that to the theta that you have in the formula.

 

[Pochen Liu]

Is this the correct diagram? (The question I am doing refers to theta from the horizontal btw)
If so, I don't understand why it is drawn like this, as in what is that 3rd vector I have labelled
I have drawn this diagram purely because I know this is the only way I can get mgsin(theta) to be along the radial direction

And in the second diagram as the ball moves towards the top of the circle theta gets smaller, but as the ball moves towards the top of the circle theta actually gets bigger.

I don't understand!

 

[jbriggs444]

Is this the correct diagram?

You add vectors head to tail, not tail to tail.

 

[nasu]

The normal force on the ball is outwards and not towards the center. Unless is zero.