Honors

Go through the lab. Make sure your errors are less than 2%. If not see me tomorrow.

Be ready to finish it quickly tomorrow, i.e. take much less than one period. Remember you need to do only through case 5c to complete the lab. For case 5 make sure you understand that one of the torques is provided by the meter stick itself. In case 5a we analyze it assuming that all the mass of the meter stick acts at the original balance point (very near 50 cm) . Does your analysis for 5a provide an acceptably small error?

(From now on we are assuming you are facing the meter stick with 0 on your left and 100 on your right)
For case b analyze it by treating the meter stick as two pieces but ignore the piece on the short i.e.left side (this is obviously a mistake so we should get a big error). The large piece has a mass = total stick mass x (100cm - the new pivot point)/100cm, i.e. that fraction to the right of the pivot. This mass acts as if it was all located at the middle of the piece of stick to the right of the new pivot point, ( this should be somewhere around the 65 cm mark or more precisely at: new pivot + 1/2 [100 -new pivot]). This will provide a clockwise torque. You will probably get a 20% or so error for 5b.

For 5c you start with 5b and then add the torque from left side of the meter stick in to the counter clockwise torque. The left side has a mass of total stick mass x (new pivot/100). The torque is as if all this mass were at the center of this piece of this part of the stick i.e., at new pivot /2. Adding this counterclockwise torque in should get you back to pretty low error.