Consider a bead of mass m sliding without friction on a wire...

Consider a bead of mass $m$ sliding without friction on a wire that is bent in the shape of a parabola and is being spun with constant angular velocity $\omega$ about its vertical axis, as shown in Figure 7.17. Use cylindrical polar coordinates and let the equation of the parabola be $z=k \rho^{2} .$ Write down the Lagrangian in terms of $\rho$ as the generalized coordinate. Find the equation of motion of the bead and determine whether there are positions of equilibrium, that is, values of $\rho$ at which the bead can remain fixed, without sliding up or down the spinning wire. Discuss the stability of any equilibrium positions you find.

Key Concepts

Lagrangian Mechanics

Lagrangian mechanics is a reformulation of classical mechanics that uses the difference between kinetic and potential energy (the Lagrangian) to derive the equations of motion through the Euler–Lagrange equation. This framework is particularly useful in systems with constraints or symmetry, as it allows one to select appropriate generalized coordinates that simplify the problem.

Generalized Coordinates

Generalized coordinates are parameters that uniquely define the configuration of a system relative to its constraints. They enable the description of the system's dynamics with a minimal number of independent variables, which simplifies the analysis of motion, especially in systems like a bead confined to move along a constrained path.

Cylindrical Polar Coordinates

Cylindrical polar coordinates (?, ?, z) are a three-dimensional coordinate system that is particularly useful for problems possessing rotational symmetry about an axis. When a system involves rotation about a vertical axis, these coordinates naturally accommodate the geometry of the problem, simplifying both kinetic and potential energy expressions.

Constraint Dynamics

Constraint dynamics involves incorporating restrictions on the motion of a system, such as the bead being confined to a parabolic path. This constraint reduces the number of degrees of freedom and can be directly incorporated into the Lagrangian, either by solving the constraint explicitly or by using methods like Lagrange multipliers.

Equilibrium and Stability Analysis

Equilibrium positions in a dynamical system are configurations where all net forces or generalized forces vanish, leading to static or constant-motion states. Stability analysis then examines whether small perturbations around these points lead to restoration (stable equilibrium) or further deviation (unstable equilibrium). This often involves checking the second derivative (or curvature) of an effective potential energy function.

Transcript

0:00 All right.

00:01 All right, we got a bead sliding on a wire.

00:06 Radius is 15 centimeters.

00:14 I've shown in a figure, 67.

00:21 It's going to take me a few pages to get there.

00:30 67.

00:33 Okay, 67.

00:36 So we've got a circle with a...

00:47 Speed on it, it's showing an angle theta, but it's also showing a rotation here.

01:00 So let's see what we got with this.

01:09 So the period of rotation is 0 .450 seconds.

01:25 Well, omega is 2 pi over the period.

01:37 And centripetal acceleration is omega squared r, which would be 4 pi squared over t squared times r.

01:56 This is r.

01:58 Okay.

02:17 Okay.

02:18 And so, using a calculator, the centripetal acceleration could be calculated.

02:39 So let's do that.

02:45 A sub c is going to be 4 pi squared over t squared, but t is 0 .45, and r is 0 .15.

03:12 Okay, back up here, over t squared times r.

03:22 Okay, so that's the centripetal acceleration.

03:28 If we draw a diagram of what's going on here, we've got gravity, and then we've got centripetal acceleration, this direction.

03:59 This is showing me that this is not.

04:02 All there is to it.

04:05 There needs to be more.

04:08 The distance down here is going to be r sine theta and we don't know.

04:35 We're trying to figure that out.

04:38 So the centripetal acceleration is going to be that times a sign of theta.

04:46 All right, let me think about that.

04:49 I'm not sure exactly what to do with that, but, um, all right.

05:00 Now, getting stray marks on this.

05:08 Okay.

05:31 Okay.

05:32 So now i drew the centripetal acceleration in the wrong direction.

05:42 Uh, so i'm going to erase here.

05:49 Now, if i draw a m .g.

05:52 At the end.

05:55 And then i draw the resultant, which would be this vector plus this vector, but erase.

06:08 I just want to draw acceleration vectors.

06:13 Then i need that to equal, i think i need this, whoops, i think i need this to be theta, or do i need it to be 90 minus theta? i want it to be perpendicular to the tangent line.

07:02 So if i draw a tangent line here, then i need that angle to be directed toward the center.

07:25 Theta, let me think about where was theta in the drawing? oh, theta's up there in the drawing.

07:38 So down here in the drawing would be 90 minus theta...

Please give Ace some feedback