Kinematics in Cylindrical Coordinates

[Zinggy]

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), v_z = 0 and the bead has an initial angular momentum L_o < 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?

 

[kuruman]

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), v_z = 0 and the bead has an initial angular momentum L_o < 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.

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?

 

[Zinggy]

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.

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.

 

[Ray Vickson]

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.

 

[kuruman]

When I'm missing gravity, does it go in the first derivative?

Have you studied Newton's Second law $\vec F_{net}=m \vec a~$?