For the particle of Prob. 12.74 , show ( a ) that the velocity of the particle and the central f

For the particle of Prob. 12.74 , show ( a ) that the...

Key Concepts

Radius of Curvature in Particle Trajectories

The radius of curvature is a geometric measure that indicates how sharply a particle's path curves at a given point. It is related to the normal (centripetal) component of the acceleration and the square of the speed. In central force problems, understanding how the radius of curvature scales with distance helps clarify the shape and behavior of the trajectory.

Scaling Relationships in Dynamics

Scaling relations involve determining how various physical quantities change with respect to one another. In the context of central force motion, it is often useful to establish proportionality between key variables such as velocity, force, and distance from the center. These relationships shed light on the system's behavior under different scales, which is critical in both theoretical analysis and practical applications.

Central Force Motion

This concept describes the motion of a particle under a force that is always directed along the line joining the particle and a fixed point (the center). This framework is fundamental in classical mechanics because it simplifies analysis by focusing on forces that depend only on the distance from the center, allowing for symmetry-based methods and direct integration of the equations of motion.

Conservation of Angular Momentum

In any central force field, the lack of a torque about the center implies that the angular momentum of the particle remains constant over time. This conservation law establishes crucial relationships between the particle's velocity and its radial distance from the center, thereby enabling the derivation of proportionality relations between these quantities.

Transcript

00:01 For this problem, we have a particle that moving under a central force.

00:06 And so the angular momentum per unit math, h, is a constant.

00:12 And so the definition of h, r squared, theta dot, must be some constant h -0, which is equivalent to its initial position r0 times its initial velocity v -0.

00:28 So if you rearrange this equation, we can find d theta d t or theta dot to be r0 v0 over r squared.

00:46 And this becomes r0 v0 over r0 squared times cost to theta.

00:59 So we're just substituting for r here.

01:01 We have an expression for r.

01:03 So we see the r knots, one r not from the top and the bottom.

01:07 Cancel.

01:08 So we get v0 over r0 times the cosine of 2 theta.

01:16 So we have an expression for theta dot in terms of theta.

01:22 So now if we differentiate r with respect to time, we get r dot.

01:29 So r dot by the chain rule is the r d theta d theta d t or just theta dot and this is equal to d by d theta of our expression for r which is r not over the square root of cost 2 theta and this is multiplied by theta dot so if we first perform this differentiation of the brackets with respect to theta and then open our theta dot which we have an expression for we get that r dot simplifies to v0 times sine 2 theta over the square root of cost 2 theta so that's a bit of mathematics now differentiating again to find r double dot we will see why we need this in later on r double dot is equal to d by d theta of r dot times theta dot so this is equal to d -teta of the expression for r dot, which is v0, sine 2 -theeta over the square root of cost 2 theta.

03:16 And if we perform this derivation, well, this is multiplied by theta dot.

03:21 We perform this differentiation with respect to theta, and then again expand theta dot with our expression above, we get our double dot to be v0 squared over r0 multiplied by two cost squared 2 theta plus sine squared 2 theta over the square root of cost 2 theta.

04:00 So now that we have our double dot, let's tackle part a.

04:06 So the first thing we need to do is find vr and v theta.

04:12 So vr is simply r dot and r dot we've calculated already.

04:22 So that's v0 sine 2 theta over the square root of cost 2 theta.

04:29 And so this becomes v0, so this is v0 sine 2 theta over the square root of cost 2 theta.

04:52 So we can bring our expression for r into this.

04:56 So we write this in terms of r to be v0, r over r0, r0 times sine 2 theta.

05:07 Remember, we have an expression for r in terms of theta.

05:11 And so we have vr.

05:15 The transverse component of the velocity v theta is equal to r theta dot.

05:22 And so this is simply v0 r over r0 times cost 2 theta.

05:31 So we have the radial and transverse components of the velocity.

05:37 So hence we can find the resultant velocity.

05:41 The resultant velocity v is equal to the square root of vr squared plus v theta squared, the square root of the sum of the squares of each component.

06:00 And so this is equal to v0 r over r0 times the square root, we substitute all our expressions in here, sine squared 2 theta plus cost squared 2 theta.

06:21 And remember, sine squared theta plus cost squared theta is simply 1...

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