TransWikia.com

How does circular motion begin in this case?

Physics Asked by Muzammil ahmed on September 5, 2021

enter image description here

In the figure, Suppose the rod system(shown black) starts moving with some angular velocity say $omega$ with the bob hanging to the string. So, after sometime the ball too rotates with the same angular velocity $omega$ that of the rod system with the help of the string’s tension(the rope making a constant angle with the vertical.I don’t understand why the bob too needs to perform circular motion. Like, it could have any arbitrary motion.

One Answer

Obviously the bob does not immediately exhibit perfect circular motion (if such a perfect circle really exists in reality). It takes time for the system to settle and achieve some kind of steady state.

As the rod is started in rotation, the bob will initially be left behind and the rope will make an angle in the tangential direction (not radial yet!). Now as the rod progresses further around its circle, the bob will be forced to swing outwards, making more of a radial angle. One way of explaining this is to say that, from the reference frame of the bob, there is a centrifugal force due to the radial acceleration that accelerates it outwards. A different way of explaining this outwards swing is to say that the rod had initially accelerated the bob tangentially, but is now turning away around the circle, so, since the ball is attached to the rod, it must swing outwards.

However, the ball does not swing straight into the path of a perfect circle, but would in reality overshoot. The tension from the string would then provide a centripetal force that is greater than necessary and the ball would accelerate back radially - decreasing the angle of the string.

Now the process would repeat: the rod would turn away (or the centrifugal force would push out the bob), and the bob would again overshoot the perfect circle again.

This would result in an oscillation with the bob teetering on each side of the perfect circle.

Eventually, due to frictional forces in the pivot of the string and air resistance, the amplitude of this oscillation would decrease and the bob would settle into a theoretically perfect circle, with angular velocity $omega$, and a constant radius.


Just for fun, to help with the visualisation of this, I made an OpenSCAD animation of the oscillations I am imagining.

animation

Source:

angle = 50 + 15 * sin($t * 360 * 7);
theta = 360 * $t;

cylinder(r=1.3, h=40);
translate([0,0,40])
rotate([0,90,theta])
cylinder(r=1.2, h=30);

color("blue")
rotate([0,0,theta])
translate([30,0,40])
rotate([0,180,0])
rotate([0,-angle,0])
cylinder(r=0.5, h=20);

color("green")
rotate(theta)
translate([30+20*sin(angle),0,40-20*cos(angle)])
sphere(2);

Correct answer by Joe Iddon on September 5, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP