# Rod sliding on a frictionless surface

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?

– user42733
Apr 7, 2014 at 16:06
• Apr 7, 2014 at 16:08
• @ParthVader They had the lower point stationary. I will edit that line. There was a same question with no answer Apr 7, 2014 at 16:10
• Apr 7, 2014 at 16:12
• So wait, this is just a rod tipping over right? Rotating a quarter of a circle?
– user42733
Apr 7, 2014 at 16:12

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).

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.

• 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. Apr 8, 2014 at 3:12
• What do you mean by "from instantaneous centre of rotation"? Do you mean the position of the center of rotation? Apr 8, 2014 at 3:19
• 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. Apr 8, 2014 at 3:20
• 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? Apr 8, 2014 at 3:36
• en.wikipedia.org/wiki/Instant_centre_of_rotation Apr 8, 2014 at 3:37

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.