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A uniform rod$(m,l)$ is standing vertically on a horizontal frictionless surface. Gravity is downwards and uniform. I give its upper end a little push and off it goes. I want to find the Normal Force(and all other variables) as a function of time.

Here's what I do :

$mg-N=ma$

Where a is downward acceleration(and the only) of centre of mass wrt ground.

Next, I think of conservation of energy. I assume it rotates $\theta$ from vertical

$mg\frac{l}{2}(1-\cos\theta)=K.E.$

I have problem writing expression of its Kinetic Energy. I have studied about instantaneous centre of rotation.

So, I can write its speed and rotational kinetic energy about it as $K.E.$? Also How do I find relation between $\omega$ (angular velocity about that instantaneous centre) and velocity of rod?

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  • $\begingroup$ Can you add the link to the question that was already answered? $\endgroup$ – user42733 Apr 7 '14 at 16:06
  • $\begingroup$ Maybe: physics.stackexchange.com/questions/98904/… $\endgroup$ – DavePhD Apr 7 '14 at 16:08
  • $\begingroup$ @ParthVader They had the lower point stationary. I will edit that line. There was a same question with no answer $\endgroup$ – evil999man Apr 7 '14 at 16:10
  • $\begingroup$ This?: physics.stackexchange.com/questions/102120/… $\endgroup$ – DavePhD Apr 7 '14 at 16:12
  • $\begingroup$ So wait, this is just a rod tipping over right? Rotating a quarter of a circle? $\endgroup$ – user42733 Apr 7 '14 at 16:12
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The translational kinetic energy is simply $\frac{1}{2} m v^2$ where $v$ is the velocity of the center of mass.

Rotational kinetic energy is $E_r = \frac{1}{2} I \omega^2$. To solve the problem, we must write the velocity of the rod as function of $\omega$ (or vice versa).

enter image description here

Consider the above image. (Note that my convention for $\theta$ is different from yours - my apologies). The red vector is the linear velocity of the rod, $v$. The orange vector is the component of this velocity which is perpendicular to the rod, $v \cos(\theta)$. From the bottom point of the rod, the linear velocity a distance $l/2$ away is $v \cos(\theta)$, so the angular velocity $\frac{d\theta}{dt} = \omega = \frac{2v \cos(\theta)}{l}$.

Using this result, we can find the rotational kinetic energy to be $E_r = \frac{1}{2} \frac{m l^2}{12} \left(\frac{2v \cos(\theta)}{l}\right)^2$. This expression (along with the other energy terms) can be used to solve a differential equation to obtain $\theta$ or $v$ as a function of time. As you found in your question, $N = mg - ma = mg - m \frac{dv}{dt}$, so you can use this relation to find the normal force as a function of time.

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  • $\begingroup$ I really want to see how we would do it from instantaneous centre of rotation. Please just tell me what to do. No need to write long explanation. $\endgroup$ – evil999man Apr 8 '14 at 3:12
  • $\begingroup$ What do you mean by "from instantaneous centre of rotation"? Do you mean the position of the center of rotation? $\endgroup$ – Shivam Sarodia Apr 8 '14 at 3:19
  • $\begingroup$ I mean the point from where only rotational motion occurs. Its coordinates can be found by intersection of normals from path of centre of mass and bottommost point. $\endgroup$ – evil999man Apr 8 '14 at 3:20
  • $\begingroup$ I might just be having a bad day, but I still don't get it. The motion of the rod is completely rotation with reference to the bottommost point, but that point is accelerating, so it's not a inertial frame of reference. Is that what you mean? $\endgroup$ – Shivam Sarodia Apr 8 '14 at 3:36
  • $\begingroup$ en.wikipedia.org/wiki/Instant_centre_of_rotation $\endgroup$ – evil999man Apr 8 '14 at 3:37
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This is what I did and I think is simple and right :

Assume $v$ linear speed of centre of mass downwards and $\omega$ angular speed around it. Use the fact the bottom point has no vertical speed to find relation between $v$ and $\omega$. And I am done.

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