3
$\begingroup$

Consider A Rigid Rod hinged at its top point Whirled around in a circle (similar to a conical pendulum). It is given that the angular velocity (and thus the semi-vertical angle) is constant. I am trying to analyse this sytem:

  1. From the rotating frame, the rod is at rest. Thus, The net torque must be zero. We have to consider Pseudo forces as well. Consider a mass element $dm$ of length dx at a distance x from the hinge point. It rotates in a circle of radius $xsin(\theta)$ so it exerts a torque: $(dm)(\omega^2)(xsin\theta)(xcos\theta)$ about the suspension point. This must be balanced by the Torque of weight=$dmgxsin\theta$. Equating Both these integrals gives a relationship between $\omega$ and $\theta$.
  2. However, from the ground frame , It is clear that there will be a Net torque (due to weight) about the suspension point, which made me question the fact that $\omega$ is constant. However, if you consider the axis of rotation,$mg$ cant produce a torque, since its parallel to the axis! So is that why $\omega$ is constant? What then , is the significance of the Net torque being non zero about the suspension point?
  3. An (unsolved) illustration in my textbook asks for the " rate of change of angular momentum" of the rod. Since this is equal to the "torque", what should I consider? Torque about the suspension point? or about the axis? and why?
$\endgroup$
3
  • 1
    $\begingroup$ You forgot to add tension of the rod $\endgroup$
    – maverick
    Apr 23, 2020 at 7:37
  • $\begingroup$ @maverick Wont that always pass through the Point of suspension and axis? It shouldnt produce a torque $\endgroup$
    – satan 29
    Apr 23, 2020 at 7:44
  • $\begingroup$ yes about suspension point there wont be torque due to tension, but it can be useful to balance forces which are in equilibruim $\endgroup$
    – maverick
    Apr 23, 2020 at 7:48

2 Answers 2

1
$\begingroup$

I think you have misunderstood the question, where the rod is not rotating about the axis, but is whirled, thus implying that it is being forced to rotate at a constant frequency, and not yet in equilibrium. Thus the assumption that the semi- vertical angle is constant is incorrect. The angular velocity is constant, but the moment of inertia about the axis changes with the angle, thus changing the angular momentum. This is probably what you have been asked to calculate (maybe instantaneous rate of change at a given angle, I don't know the exact question). You can do this using a free body diagram.

$\endgroup$
1
$\begingroup$

Typically we try to find a relation between the opening angle of the cone (between the rod and the vertical) and the angular speed of rotation.

It is decided to study the problem in the inertial frame of reference given by point O where the bar is tied to the ceiling. The motion of the rod is a rotation around the vertical axis passing through O, which is not an axis of symmetry of the rod. Not being an axis of symmetry, the angular momentum of the rod with respect to the point O is not parallel to the angular velocity and follows a precession motion. The angular momentum precedes with the same speed $\vec\omega$ with which the rod rotates and therefore its motion follows the law: $$ {d\vec L_o \over dt} = \vec \omega \times \vec L_o $$ For the second cardinal equation, the derivative of the angular momentum over time with respect to a fixed pole is equal to the torque. The forces involved in this inertial system are exclusively the constraining force of the ceiling and the weight force. The constraining force of the ceiling produces no torque as it is applied at point O. The weight force is actually a whole field of parallel weight forces acting on each point of the object. We apply the parallel force field theorem which tells us that the total torque given by the weight forces is equal to the torque of the total weight force applied in the center of gravity, in the middle of the bar. $$ {d\vec L_o \over dt} = \vec \omega \times \vec L_o = \vec \tau_o = \vec r_{CM} \times \vec W $$ We observe that both vector products actually give rise to a vector in the outgoing direction. We impose equality between the modules $$ \vec \omega \times \vec L_o = r_{CM} \times \vec W $$ $$ \omega L_o \sin(\pi/2 - \theta) = {l \over 2} Mg \sin(\theta) $$ $$ \omega L_o \cos(\theta) = {l \over 2}Mg\sin(\theta) $$ $$ \omega L_o = {lMg \over 2}\tan(\theta) $$ The angular momentum $ L_o $ seems difficult to calculate because the object rotates around a non-symmetrical axis. However, we can calculate it simply from the definition: $$ \vec L_o = \sum_i \vec r_i \times m_i\vec v_i $$ $$ \vec L_o = \sum_i m_i\vec r_i \times (\vec \omega \times \vec r_i) $$ The summation takes vectors with the same direction and orientation therefore the module of the summation is the sum of the modules. By paying attention in the calculation of the modules we get $$ L_o = \sum_i m_ir_i^2w \sin(\pi - \theta) = \sum_i m_ir_i^2w \sin(\theta) $$ The rod has a continuous and uniform distribution of matter so we integrate along its length. $$ L_o =\int_{0}^{l} r^2\omega\sin(\theta)dm =\int_{0}^{l} r^2\omega\sin(\theta){M\over l} dr = \omega\sin(\theta){M}{l^2 \over 3} $$ Substituting in the equation found previously: $$ \omega^2\sin(\theta){M}{l^{{2}} \over 3} = {{lM}g \over 2}\tan(\theta) $$ $$ \omega^2 = {3g \over 2l\cos(\theta)} $$ Which expresses the relationship between the angular velocity and the angle between the rod and the vertical. From this expression it is also possible to derive a minimum speed of rotation for the conical pendulum regime to be sustained: $0<\cos(\theta) < 1$ so $\omega^2_{\text{min}} = 3g/2l$.

The same results can be obtained by placing ourselves in a non-inertial reference frame rotating with the rod and centered in O. In this case the apparent centrifugal force, different from point to point of the rod, and the weight force generate the torque that allows rotation. In the chosen system the object is stationary and therefore it can be imposed that the sum of the external torques is zero. By placing oneself in the non-inertial system, one can therefore make the problem a statics problem.

This is my take on the problem. Trying with the non-inertial frame I got the same result.

$\endgroup$

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.