A ring is fitted loosely (with no friction) around a long,...

A ring is fitted loosely (with no friction) around a long, smooth rod of length $L=0.50 \mathrm{~m} .$ The rod is fixed at one end, and the other end is spun in a horizontal circle at a constant angular velocity of $\omega=4.0 \mathrm{rad} / \mathrm{s} .$ The ring has zero radial velocity at its initial position, a distance of $r_{0}=0.30 \mathrm{~m}$ from the fixed end. Determine the radial velocity of the ring as it reaches the moving end of the rod.

Key Concepts

Motion in Polar Coordinates

In dynamics, using polar coordinates is essential for analyzing situations where motion naturally splits into radial and angular components. This framework involves expressing a particle's velocity and acceleration as combinations of changes in the radius (radial component) and in the angle (tangential component). It is particularly useful in problems where the path or force field has circular symmetry, allowing for a direct connection between the geometric constraints and the dynamic behavior of the system.

Dynamics of Constrained Motion in Rotational Systems

When a particle is forced to remain in contact with a rotating object, such as a bead or ring on a rod, its motion is constrained so that its angular position follows that of the rotating object. This constraint typically fixes the angular velocity while allowing the radial coordinate to vary. Analyzing the radial dynamics leads to differential equations—often with constant coefficients—that can yield solutions in terms of hyperbolic functions. Understanding these systems involves applying Newton’s laws with the recognition that constraint forces (which do work) shape the particle’s evolution in the constrained direction.

Transcript

00:01 Hello and welcome to this video solution of numerate here we are given that there is a ring that is fitted loosely with no friction around a long smooth rod of length l so length of the rod is 0 .5 meters the rod is fixed at one end and the other end is spun in a horizontal circle at constant angular speed of omega which is 4 radians per second, right? the ring has a zero radial velocity at its initial position, which is r naught which is at 0 .3 meters now you have to determine the radial velocity as the ring reaches the moving end of the rod right, so here what you can do is you have to calculate the accelerations, right? so here we can say that the radial acceleration of the ring right will be equal to the centripetal acceleration itself, right? ac and here you can write the radial acceleration is dv dt that is equal to omega square r right r is the distance from the center right and here you can have like dv dr over dr dt, right you can write like this chain rule.

01:30 This is omega square r and dr dt is nothing but vr right dv dr that is equal to omega square r take it as only v here right and you can write v dv that is equal to omega square r dr right, you can take integral on both sides now and this will be from initial velocity u to v let's say and here you will be having r naught to l as a limit of the radius right and here you can have like half this is v square minus u square that is equal to half omega square l square minus r square r naught square, right? this is what you have and the final velocity v that's equal to omega root over of l square minus r naught square right you can take initial speed u to be 0 this u to be 0 or here you can write 0 itself and you can do it 0 square here and you can finally plug in the value this is 4 omega is 4 and the length is given to be 0 .5 minus this is 0 .3 squared right and this comes out to be around 1 .6 right meters per second as a velocity...

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