Kinematics in Cylindrical Coordinates

In summary: The Attempt at a Solution This is the diagram that goes with the question.The bead is sliding along the outside of the cylinder. When I'm missing gravity, does it go in the first derivative? The position vector is given without gravity present.Thanks for your time. Gravity affects acceleration, not velocity.
  • #1
Zinggy
12
0

Homework Statement



A small bead of mass m slides on a frictionless cylinder of radius R which lies with its cylindrical axis horizontal. At t = 0 , when the bead is at (R,0), vz = 0 and the bead has an initial angular momentum Lo < mR sqrt(Rg) about the axis of the cylinder where g is the acceleration due to gravity. The bead slides from (R,0) down the curved surface of the cylinder and eventually loses contact with that surface.

Find r(double dot) in Cylindrical Coordinates.
2. Homework Equations

Position Vector of r = ˆ iRcosφ + ˆ jRsinφ = ρˆR

The Attempt at a Solution


I know r(double dot) is the same as d2r/dt2 so;
r = ˆ iRcosφ + ˆ jRsinφ = ρˆR
r(dot) =- iRsinφ+jRcosφ -unless R changes with time, which I don't think it does.
then
r(double dot) = iR-cosφ - jRsinφ
I think I'm missing something. Do I need to include the angular momentum somewhere?
 
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  • #2
Zinggy said:

Homework Statement



A small bead of mass m slides on a frictionless cylinder of radius R which lies with its cylindrical axis horizontal. At t = 0 , when the bead is at (R,0), vz = 0 and the bead has an initial angular momentum Lo < mR sqrt(Rg) about the axis of the cylinder where g is the acceleration due to gravity. The bead slides from (R,0) down the curved surface of the cylinder and eventually loses contact with that surface.

Find r(double dot) in Cylindrical Coordinates.
2. Homework Equations

Position Vector of r = ˆ iRcosφ + ˆ jRsinφ = ρˆR

The Attempt at a Solution


I know r(double dot) is the same as d2r/dt2 so;
r = ˆ iRcosφ + ˆ jRsinφ = ρˆR
r(dot) =- iRsinφ+jRcosφ -unless R changes with time, which I don't think it does.
then
r(double dot) = iR-cosφ - jRsinφ
I think I'm missing something. Do I need to include the angular momentum somewhere?
You are missing the force of gravity in your equations.

Zinggy said:
r(dot) =- iRsinφ+jRcosφ -unless R changes with time, which I don't think it does.
Your are also missing the chain rule when you take derivatives with respect to time.
Is the bead on the inside or the outside of the cylinder? Specifically, is point (R,0) in Cartesian coordinates?
 
  • #3
kuruman said:
You are missing the force of gravity in your equations.Your are also missing the chain rule when you take derivatives with respect to time.
Is the bead on the inside or the outside of the cylinder? Specifically, is point (R,0) in Cartesian coordinates?

This is the diagram that goes with the question.
Physics diagram.png

The bead is sliding along the outside of the cylinder. When I'm missing gravity, does it go in the first derivative? The position vector is given without gravity present.

Thanks for your time.
 

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  • Physics diagram.png
    Physics diagram.png
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  • #4
Zinggy said:
This is the diagram that goes with the question.
View attachment 238105
The bead is sliding along the outside of the cylinder. When I'm missing gravity, does it go in the first derivative? The position vector is given without gravity present.

Thanks for your time.
Gravity affects acceleration, not velocity.

You must resolve the gravitational force into two components: (i) the radial component (pointing towards the center of the circle); and (ii) the tangential component (pointing southwest in your drawing, along the surface of the cylinder). The radial component just serves to hold the bead to the surface, while the tangential components accelerates the bead along the surface.
 
  • #5
Zinggy said:
When I'm missing gravity, does it go in the first derivative?
Have you studied Newton's Second law ##\vec F_{net}=m \vec a~##?
 

Related to Kinematics in Cylindrical Coordinates

1. What is kinematics in cylindrical coordinates?

Kinematics in cylindrical coordinates is a branch of physics that deals with the motion of objects in a three-dimensional cylindrical coordinate system. It involves the study of position, velocity, and acceleration of objects moving along curved paths in cylindrical coordinates.

2. How is kinematics in cylindrical coordinates different from kinematics in Cartesian coordinates?

Kinematics in cylindrical coordinates differs from kinematics in Cartesian coordinates in that it takes into account the cylindrical nature of the coordinate system. In Cartesian coordinates, motion is described in terms of x, y, and z coordinates, while in cylindrical coordinates, it is described in terms of radius, angle, and height.

3. What are the advantages of using cylindrical coordinates in kinematics?

One advantage of using cylindrical coordinates in kinematics is that it simplifies the equations of motion for objects moving in circular or cylindrical paths. It also allows for a more intuitive understanding of rotational motion and makes it easier to solve problems involving objects moving in cylindrical coordinates.

4. How do you convert between Cartesian and cylindrical coordinates in kinematics?

To convert between Cartesian and cylindrical coordinates in kinematics, you can use the following equations:

x = r cosθ

y = r sinθ

z = z

where r is the distance from the origin, θ is the angle from the positive x-axis, and z is the height.

5. Can kinematics in cylindrical coordinates be applied to real-world situations?

Yes, kinematics in cylindrical coordinates can be applied to real-world situations. It is commonly used in fields such as engineering, physics, and astronomy to describe the motion of objects in cylindrical systems, such as rotating machinery, celestial bodies, and fluid flow in pipes.

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