0
$\begingroup$

I have started learning about rotating torque from Berkely Physics Course-1. I came across the bead on massless rod setup as shown in the figure below.enter image description here

Since omega and angular momentum are not aligned, bearings must be added for sustained rotation in the above configuration.

What I don't understand, is what is providing the opposite torque in absence of the bearings.(i.e. making the rod vertical).By making vertical I mean, making theta=π/2. I, guess it is the tension, but how can tension provide torque, since it's directed towards origin.

Improve this question
$\endgroup$
4
  • $\begingroup$ Are you asking what happens when $\theta = \pi/2$ ? $\endgroup$
    – John Alexiou
    Mar 20, 2019 at 14:30
  • $\begingroup$ @ja72 umm no. In that case tension provides the centrepetal force. But what is the role of tension in general case(i.e theta not equal to 0) $\endgroup$
    – user215736
    Mar 20, 2019 at 15:43
  • $\begingroup$ I guess I got confused by your statement "making the rod vertical". It is possible to edit the question and clarify the scenario you are asking about. $\endgroup$
    – John Alexiou
    Mar 20, 2019 at 16:34
  • $\begingroup$ @ja72 done. I mean if we don't add bearings the rod quickly becomes vertical. How is that happening $\endgroup$
    – user215736
    Mar 20, 2019 at 16:37

1 Answer 1

Reset to default
1
$\begingroup$

The statement in your last comment

I mean if we don’t add bearings the rod quickly becomes vertical.

is not correct.

If the bearings were removed at the instant which is shown in the diagram the angular momentum $\vec J$ will not change as there are no external torques acting on this system of two masses and the rod.

What you would see is the two masses rotating at constant speed on opposite sides of a circular trajectory of radius $a$ whose plane is perpendicular to the angular momentum vector with the tension forces causing the centripetal acceleration of the masses.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah now I understand. Thanks. I was having a very wrong assumption. $\endgroup$
    – user215736
    Mar 20, 2019 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.