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Relating $v$ and $\omega$

2 votes

A rod of length $l$ is standing vertically on a frictionless surface.
It is disturbed slightly from this position. Let $\omega$ and $\alpha$
be the angular speed and angular acceleration of the rod when the rod
turns through an angle $\theta$ with the vertical, then find the value
of the acceleration of the centre of mass of rod.


Using conservation of energy,

$\dfrac 12 mv^2 + \dfrac 1 2 I_{cm}\omega^2 = \dfrac{mlg}{2}(1-\cos \theta)$

$I_{cm}= \dfrac {ml^2}{12}$

Substituting this and then differentiating wrt time,
we get:

$2a v = g\omega\sin \theta - \dfrac{2}{3}l \omega \alpha$ \where $a=$ acceleration.

Now, somehow I have to eliminate $v$. I am unable to do that. How do I proceed?

The answer given is:

$a= \dfrac l 2 \sin\theta \alpha + \dfrac l 2 \omega^2 \cos \theta $

asked Mar 23, 2018 in Physics Problems by Reststack (422 points)

1 Answer

1 vote

Notice that $g$ does not appear in the answer. This is not an equation of motion but an equation of constraint like $v=r\omega$ or $a=r\alpha$.

When the angular velocity of the rod is $\omega$ the velocity of the CM is $v'=\frac12 l \omega$ relative to the point of contact with the ground, which is itself moving horizontally . Relative to the stationary ground, the vertical component of this relative velocity is $v=v'\sin\theta$. The horizontal component of the CM velocity in the ground frame is $0$ because there are no horizontal forces on the rod so the CM falls vertically down. The lowest point of the rod has horizontal velocity $-v'\cos\theta$ along the ground. See comments in A stick is positioned vertically and allowed to fall. What is its speed?

So we have by differentiation :
$$v=\frac12 l\omega\sin\theta$$ $$a=\dot v=\frac12 l\alpha\sin\theta+\frac12 l\omega^2\cos\theta$$

To find an equation of motion, substitute for $v$ and $a=\dot v = \ddot y_C$ from the above equations, as done by ja72 in Equation of motion for a falling rod (with one end touching a frictionless surface). Then you have a 2nd order non-linear equation relating $\theta, \dot\theta, \ddot\theta$. If you are given initial values of $\theta, \dot\theta$ then you can solve this differential equation - although perhaps there is no analytical solution, only a numerical solution.

answered Mar 24, 2018 by sammy gerbil (28,448 points)
edited Apr 5, 2018 by sammy gerbil
Why is the velocity of the CM $1/2 l\omega$? Also, if it doesn't have any horizntal components then shoudn't it be *entirely* vertical velocity i,e. the vertical component should be $0.5 l\omega$ instead of $0.5 l \omega \sin \theta$?
No, for the reasons given in my 2nd paragraph. The velocity of the CM in the ground frame is vertical only, and depends upon the angular velocity $\omega=\dot\theta$ of the rod and also on its inclination $\theta$ to the vertical. When the rod is vertical the CM is stationary (velocity zero), it does not fall at all, even if $\omega$ is large. When the rod is horizontal the velocity of the CM is $\frac12 l \omega$ vertically down. Your expression $\frac12 l \omega$ does not depend on $\theta$ so it must be wrong.
Can we discuss this in the chat room? Can you give me an appointment as per your convenience?
We don't have a chatroom on this site. Do you mean the chatroom on Physics Stack Exchange? If so, I haven't found you there. Are you using a different name there? What is your question?