- 1 Place A Mass Mi 0 200 Kg At The 0 100m Position Slide The Meter Stick On Top Of The Knife Edge Object Until The To 1 (97.83 KiB) Viewed 43 times
- 1 Place A Mass Mi 0 200 Kg At The 0 100m Position Slide The Meter Stick On Top Of The Knife Edge Object Until The To 2 (179.79 KiB) Viewed 43 times
- 1 Place A Mass Mi 0 200 Kg At The 0 100m Position Slide The Meter Stick On Top Of The Knife Edge Object Until The To 3 (77.88 KiB) Viewed 43 times
1. Place a mass mi = 0.200 kg at the 0.100m position. Slide the meter stick on top of the knife-edge object until the torque exerted by mig is balanced by the torque of the meter stick weight acting at Xg. Repeat the entire process until the best balance is achieved. The position at which the meter stick is supported is Xo. Record Xo in Data and Calculations Table 3.
- = - 2. The values of the lever arms are given by dı = \xı – xol and do = |xg – xol. Calculate and record the values of d, and do in Data and Calculations Table 3. 3. For these conditions, Etccw = migdį and Etcw = mogd, where where mo = Στε» stands for the assumed unknown mass of the meter stick. Equating the two torques gives mo = mi(di/d.). Calculate and record in Data and Calculations Table 3 this value as (m.)exp. 4. Calculate and record in Data and Calculations Table 3 the percentage error in (mo)exp compared to the meter stick mass determined by the laboratory balance. 5. Repeat steps 1 to 4 for trial 2.
0.110 kg pivot at 0.4m Meter stick mass using laboratory balance (mo) true Position (m) Lever arm (m) (mo)exp % Error Mass (kg) X1 = 0.100 di = 0.300 do = 0.100 m mi = 0.050 d2 = m2 = 0.010 X2 = 0.800