Our goal in this lab is to derive a function of max height of a meterstick swing throught the vertical after an inelastic collision (in this case, with a piece of clay).
Our meterstick had a hole drilled at 0.2 meters from its end, which was mounted on an axis so that it would swing through the vertical. At the bottom of its path, a piece of clay is taped in place which will collide and stick to the meterstick as it passes its lowest point. In order to collect data, we filmed the process and used loggerpro to chart its position with respect to time. A screenshot of this process is below
In our trial, we found our max height to be 0.157 m.
In order to verify this, we now had to derive a theoretical maximum height. The motion of this rod has 3 parts.
First as it falls from its initial height to the bottom of its swing, gravitational PE is being converted to rotation KE.
In the second part, we have an inelastic collision or
and finally, we have rotational KE being converted to gravitational potential energy
As we can see, our final height is theoretically 0.131. Which is about 16 percent off of our experimental value.
This is a large discrepency, but unfortunately we had no time to go back and re-do the experement. We suspect that this error is due to a scaling issue when we plotted the data from our video.
Our meterstick had a hole drilled at 0.2 meters from its end, which was mounted on an axis so that it would swing through the vertical. At the bottom of its path, a piece of clay is taped in place which will collide and stick to the meterstick as it passes its lowest point. In order to collect data, we filmed the process and used loggerpro to chart its position with respect to time. A screenshot of this process is below
In our trial, we found our max height to be 0.157 m.
In order to verify this, we now had to derive a theoretical maximum height. The motion of this rod has 3 parts.
First as it falls from its initial height to the bottom of its swing, gravitational PE is being converted to rotation KE.
Mgh=(1/2)I(omega)^2
In the second part, we have an inelastic collision or
I(omega)i=I(meterstick+clay)(omega)f
and finally, we have rotational KE being converted to gravitational potential energy
(1/2)I(meterstick+clay)(omega)^2=Mgh
The inertia of the meterstick had to be determined using the parallel axis theorem,
I=Icm+M(d)^2
Our calculations are shown below
This is a large discrepency, but unfortunately we had no time to go back and re-do the experement. We suspect that this error is due to a scaling issue when we plotted the data from our video.
What I don't get is that if the pivot was at the 20 cm mark then the moment of inertia of the bar would be 1/12ML^2 + M(0.3L)^2. It is very hard to tell from what is on the piece of paper, but it looks like something else.
ReplyDeleteAlso, the ∆h of the rod's cm is different than the ∆h of the clay's cm.
ReplyDelete