# Block sliding on the inside of a cone

• m_state724
In summary, the third year physics major is trying to decide if he should continue his major because he no longer understands the concepts in the class. He came across a problem that he is unable to solve and he is looking for help. He has done a free body diagram and knows how to post it, but he does not know how to solve the equation of motion. He has reduced the problem to a single position variable and is looking for another position angle to solve for the block.
m_state724
Hello there, this is my first posting on this board. I am a third year physics major and I've taken a lot of courses but I've always just scraped by with a large amount of help from small groups. I'm thinking about reconsidering my major because I feel like I no longer can understand the concepts.. but anyways, I came across this problem and I am stumped.

Using {r, phi, z} coordinates write the equation of motion for a small block that is free to move on the inside surface of an upper cone under the influence of gravity. Repeat this problem with the addition of simple friction.

I have done the free body diagram but I don't know how to post that on here. Anways this is what i got:

r^=-Ncos(phi)=-mv^2/r
z^=Nsin(phi)-mg=0

Then divide z^ by r^ and get this:
tan(phi)=rg/(v^2)

?

I really have no idea what I am doing. Any help or sense of direction would be greatly appreciated.

Last edited:
What's an upper cone?

It's an inverted cone, it looks like this:

\/

m_state724 said:
Hello there, this is my first posting on this board. I am a third year physics major and I've taken a lot of courses but I've always just scraped by with a large amount of help from small groups. I'm thinking about reconsidering my major because I feel like I no longer can understand the concepts.. but anyways, I came across this problem and I am stumped.

Using {r, phi, z} coordinates write the equation of motion for a small block that is free to move on the inside surface of an upper cone under the influence of gravity. Repeat this problem with the addition of simple friction.

I have done the free body diagram but I don't know how to post that on here. Anways this is what i got:

r^=-Ncos(phi)=-mv^2/r
z^=Nsin(phi)-mg=0

Then divide z^ by r^ and get this:
tan(phi)=rg/(v^2)

?

I really have no idea what I am doing. Any help or sense of direction would be greatly appreciated.
It looks like you are using phi as the angle of the cone. Is that what is intended? I would think there has to be an azimuthal angle in your coordinate system, and some people use phi for that angle. Does your text or prof use it that way?. The angle of the cone establishes a fixed ratio for r and z (taking the origin to be the vertex of the cone). I take it the by r^ you mean the force in the r direction and z^ is the force in the z direction. There has to be something about the azimuthal direction in the equations of motion. Unless the mass moves only in a vertical plane (two dimensional problem of an inclined plane) the azimuthal motion is an important component.

Let's make sure the variables are correctly identified, and then take it from there.

Okay i'll try to explain the setup I did as well as I can

\|/_
the | between the \/ represents the Z axis.
The underscore represents the R axis
and the angle between |/ is Phi.

Does that help explain my setup at all? I'm trying to make it as clear as possible..

m_state724 said:
Okay i'll try to explain the setup I did as well as I can

\|/_
the | between the \/ represents the Z axis.
The underscore represents the R axis
and the angle between |/ is Phi.

Does that help explain my setup at all? I'm trying to make it as clear as possible..
Then if the vertex of the cone is at the origin, phi is just a parameter of the cone such that tan(phi) = r/z = constant. How are you going to describe the position of the block in three dimensions when you only have two position coordinates (r and z) that have a fixed ratio? Your problem has been reduced to two dimensions with a constraint that reduces it to a single independent position variable. The real problem is a three dimensional problem with a constraint that relates two of the variables, reducing the number of independent variable to two. You need another position angle in the problem.

\|/_
the | between the \/ represents the Z axis.
The underscore represents the R axis
and the angle between |/ is Alpha, Like you said tan(alpha)=r/z.
Phi is the angle of where the block is located inside of the cone, and it can be integrated from 0 to 2Pi to find the location of the block.

Does that sound better?

m_state724 said:
\|/_
the | between the \/ represents the Z axis.
The underscore represents the R axis
and the angle between |/ is Alpha, Like you said tan(alpha)=r/z.
Phi is the angle of where the block is located inside of the cone, and it can be integrated from 0 to 2Pi to find the location of the block.

Does that sound better?
I would not use the word "integrated" but I know what you mean, and yes this sounds better. Phi is the azimuthal angle coordinate. Now try writing your equations of motion including the motion in the azimuthal direction.

\|/_
the | between the \/ represents the Z axis.
The underscore represents the R axis
and the angle between |/ is Alpha, Like you said tan(alpha)=r/z.
Phi is the angle of where the block is located inside of the cone, and it can be integrated from 0 to 2Pi to find the location of the block.

phi^=[m(v^2)]/r
r^=-Ncos(alpha)
z^=Nsin(alpha)-mg=0
F=phi^ + r^ + z^

Thanks for sticking with me. I know this is elementary to most everyone, I've taken E&M I and II, Modern, Thermo, and now I'm in classical and I feel like I've lost a grip on the fundamentals.

m_state724 said:
\|/_
the | between the \/ represents the Z axis.
The underscore represents the R axis
and the angle between |/ is Alpha, Like you said tan(alpha)=r/z.
Phi is the angle of where the block is located inside of the cone, and it can be integrated from 0 to 2Pi to find the location of the block.

phi^=[m(v^2)]/r
r^=-Ncos(alpha)
z^=Nsin(alpha)-mg=0
F=phi^ + r^ + z^

Thanks for sticking with me. I know this is elementary to most everyone, I've taken E&M I and II, Modern, Thermo, and now I'm in classical and I feel like I've lost a grip on the fundamentals.

In the absence of friction, there is no force in the phi direction. The normal force is the only force acting other than gravity. The radial force is the centripetal force, as you originally had it, but you want to express it in terms of angular velocity rω = rdφ/dt, v^2 = r^2ω^2. Any velocity components in the r or z direction do not contribute to the centripetal acceleration.

## 1. How does the radius of the cone affect the speed of the block?

The radius of the cone has a direct impact on the speed of the block. The smaller the radius, the faster the block will slide down due to the shorter distance it needs to cover. On the other hand, a larger radius will result in a slower speed as the block has to cover a greater distance.

## 2. What is the role of friction in block sliding on the inside of a cone?

Friction plays a crucial role in determining the speed and movement of the block inside the cone. It provides the necessary force for the block to move and also acts as a resistance force, slowing down the block's speed. Without friction, the block would not be able to slide down the cone.

## 3. Can the angle of the cone affect the block's trajectory?

Yes, the angle of the cone can affect the trajectory of the block. A steeper angle will result in a faster and more direct descent, while a shallower angle will cause the block to slide in a curved path. The angle also affects the amount of friction acting on the block and, therefore, its speed.

## 4. Is the mass of the block a significant factor in block sliding on the inside of a cone?

The mass of the block does play a role in block sliding on the inside of a cone, but it is not as significant as other factors like the angle and radius of the cone. A heavier block may slide slightly faster due to its greater weight, but the difference in speed will not be significant.

## 5. How is energy conserved in block sliding on the inside of a cone?

In a frictionless system, the total energy of the block remains constant throughout its descent. As the block moves down the cone, potential energy is converted into kinetic energy, and vice versa. However, in a real-world scenario with friction, some energy will be lost as heat due to the work done by friction. Therefore, energy is not completely conserved in this system.

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