What Are the Dynamics of a Particle in Vertical Circle Motion?

[rulo1992]

Homework Statement

Suppose a non-uniform circular motion where a particle of mass "m" is attached to a string, which rotates on a vertical plane. Once an initial velocity is provided to the particle at the lowest point of the trajectory, no further forces act on the particle. (Air drag is negligible)

1. Which is the minimum velocity that the particle requires to reach the highest point of the trajectory?
2. If initial velocity is two-times the one we calculated in question 1, what would be the velocity in the highest point? Calculate the acceleration too:
3. Calculate the maximum height the particle can reach if the velocity is half the one calculated in 1.
4. Calculate the initial velocity if the particle rotates only 2$\Pi$/3 radians:

Homework Equations

$1/2$mv^2_bottom=$1/2$mv^2_top+mg2r
Centripetal acceleration=(v^2)/(r)
v=ωr

The Attempt at a Solution

For no.1 , given that KE is fully transformed into potential energy at the highest point of the circle,KE is 0 in the right side of the equation so: v=2√(gr)

For number 2, if velocity is 4√(gr) then $1/2$m(4√(gr))^2=$1/2$mv_top^2+mg2r →v_top=2√(3gr)

So I tried to calculate the centripetal acceleration and I got a_n=(2√(3gr)/r
But know I don't know how to calculate the tangential acceleration with only the values I know and now I'm stuck on this problem. Pleasse help me!

 

[haruspex]

For no.1 , given that KE is fully transformed into potential energy at the highest point of the circle!

If all KE converted to PE, how is it still moving in a circle?

 

[ehild]

The speed is minimum at the top of the circle. How is tangential acceleration related to speed? ehild

 

[rulo1992]

If all KE converted to PE, how is it still moving in a circle?

Well question no. 1 asks for the minimum velocity required to get to the top of the circle, so the particle does not necessarily move in a cirlcle after that point.

 

[ehild]

Well question no. 1 asks for the minimum velocity required to get to the top of the circle, so the particle does not necessarily move in a cirlcle after that point.

It will not even reach that point along the circle as the string becomes slack before.

ehild

 

[rulo1992]

The speed is minimum at the top of the circle. How is tangential acceleration related to speed?


ehild

If the velocity vector(tangent to the trajectory) is antiparallel to the tangential acceleration vector, speed decreases.
But that doesn't answer my question :(
Sorry but this is my first post and I don't know how to use latex

 

[ehild]

Well, an other hint: What is the direction of the resultant force at the top of the circle? And how are acceleration and force related?

ehild

 

[rulo1992]

Well, an other hint: What is the direction of the resultant force at the top of the circle? And how are acceleration and force related?

ehild

Ohh I get it the resultant force points towards the center of the circle, therefore there's no tangencial acceleration right?

 

[ehild]

It is right!

ehild

 

[rulo1992]

Hi, I have another question, in another section of my homework it asks me to calculate the initial velocity if the object traveled to 2pi/3 radians, but how can I do it if I don't know any angular velocity nor acceleration?

 

[ehild]

Apply conservation of energy again.

ehild

 

[rulo1992]

Can I use polar coordinates? so y=rsen120 when the angle is 2pi/3 radians, and so the height at that point would be r+rsen120?

 

[ehild]

Can I use polar coordinates? so y=rsen120 when the angle is 2pi/3 radians, and so the height at that point would be r+rsen120?

Yes, the height is R(1+sin(120°)).

Edit: Oops. It is R(1-cos(120°))

ehild

 

[rulo1992]

And considering the circle rotates counterclockwise and at 2∏/3 radians tangential acceleration exists again(parallel to velocity, so it accelerates) , is it really possible for the object to only rotate to 2∏/3?

 

[ehild]

The particle moves along the circle. See picture. The height from the lowest point is h=R(1-cos(θ)) .
It does not matter if theta is measured clockwise or anti-clockwise as cosine is even function.

ehild

 

[rulo1992]

Oh yes sorry my mistake we have to use cos indeed. Another question if you don't mind, will this initial velocity depend on the tension of the string?

 

[ehild]

The initial velocity is given. The tension of the string depends on the speed. It has to provide the necessary centripetal force together with the radial component of the weight.

ehild

 

[rulo1992]

What I don't get is how is it possible to calculate the initial velocity using conservation of energy.
Because if height=(R(1-costheta)) , is it correct to use the formula:1/2m(velocityatbottom)^2=mg(R(1-costheta))? or am I doing something wrong?
I don't get how can kinetic energy be 0, if the particle keeps moving along the cirlce after that point

 

[ehild]

The problem text said that

Calculate the initial velocity if the particle rotates only 2Π/3 radians

So the particle does not go around its circular path after covering 2pi/3 radians.

If he particle is fixed to a rigid rotating stick that means the particle stops at that angle, then starts to move backwards.

If it is connected to a string, it still can have speed after reaching that angle, but the radial component of the weight is not enough for the centripetal force, the string gets slack, the motion will not be circular any more.
I think the problem maker expect the solution according to the first case.

ehild

 

[rulo1992]

How are Newton's 1st and 3rd laws applied to the forces acting on the particle?
this is the last question of my homework, and I'm having problem answering it correctly in a few lines. Do I have to consider inertial and non inertial frames separately?

 

[ehild]

Stay in the inertial frame of reference. Consider the forces acting on the particle: the tension in the string and gravity. The radial component of the resultant is the centripetal force, while the tangential component accelerates the particle along the circle.

When discussing Newton's 3rd law, be careful. You have two kinds of interaction. One between the particle and the string. The strings pulls the particle with its tension. The particle pulls the string and that causes the tension in the string.
The other interaction is the gravitational pull between Earth and particle.

ehild

 

[olivermsun]

What I don't get is how is it possible to calculate the initial velocity using conservation of energy.
Because if height=(R(1-costheta)) , is it correct to use the formula:1/2m(velocityatbottom)^2=mg(R(1-costheta))? or am I doing something wrong?

Well you know that the height at the top of the circle is just $2R$ (you can check that $\theta = 180^\circ$ and $\cos 180^\circ = -1$), so that part is fine.

I don't get how can kinetic energy be 0, if the particle keeps moving along the cirlce after that point

As long as KE > 0 at the highest point in the circle, the motion will continue, so what you do find the velocity $v_\mathrm{min}$ that gives exactly KE = 0. Then you know that the initial velocity $v_0$ has to satisfy $v_0 \gneq v_\mathrm{min}$.

 

[haruspex]

As long as KE > 0 at the highest point in the circle, the motion will continue,

This is wrong. There is a minimum speed at the top of the circle, and it is greater than zero. It won't reach the top if the string goes slack at any time before that.
What forces act on it at the top? What acceleration does it have there?

 

[olivermsun]

This is wrong. There is a minimum speed at the top of the circle, and it is greater than zero. It won't reach the top if the string goes slack at any time before that.

Yup you're right. Thanks for catching that.

v^2/R = g is the limiting condition, where gravity is exactly equal to the centripetal force needed to follow the circle (and hence the string need not provide any tension). Does that sound right to you?

 

[rulo1992]

how can I express the forces that act on the particle with x, y coordinates instead of polar coordinates?

 

[olivermsun]

Well the good thing is that g is always down (pointing in the -y direction).

The force from the string is inward radial, so you will have to decompose it into x, y components using sines and cosines.

 

[rulo1992]

If the initial velocity is not enough to get to the top, would the movement then become a simple-harmonic motion?

 

[haruspex]

If the initial velocity is not enough to get to the top, would the movement then become a simple-harmonic motion?

No. A very small movement approximates SHM. A movement beyond the horizontal position is quite different again, since the string may become slack and the motion is a free fall trajectory (parabola) for a while.
The question you need to answer is what speed the object needs to be traveling at at the top of the circle in order to keep the string just taut. Since it will be only just taut, you can take the tension as effectively zero. What forces therefore act on it? What acceleration does that result in? If it is traveling at speed v in a circle radius r, what is its actual acceleration?