Question
In the case shown above, we have: tau_1 = m_1gl_1, tau_2 = m_2gl_2, tau_2 = m_2gl_2. It should be noted that torque is a vector (), and as such, the direction of the torque must be accounted for; for example, a clockwise rotation yields a negative torque value. Meter stick Meter stick clamps (knife edge) Weights Fulcrum (balance support) Data collection Balance and secure the meter stick in the knife edge clamp. Record the position of the balance point (center of mass) of the meter stick relative to the left end of the stick and as I_0. Using the clamps, hang m_1 = 50g on the right end of the meter stick and hang m_2 = 100g on the other side. Do you need to include the mass of the clamp? Adjust the position of the masses by sliding them along the ruler until the system is balanced. Record these distances from the pivot point in Table 1. Place the masses used above on the left side of the balance point, one at 20 cm and the other at 30 cm, now add a third mass on the right side (as in Figure 1). Slide the third mass around until the system balances. Record the masses and lever arms in Table 2. Calculations For Experiment 1, find the net torque of the system: What is the theoretical lever arm for m_2 for Experiment 1? Use this value to find the percent error in the net torque. For Experiment 2, find the net torque of the system: What is the theoretical lever arm for the mass on the right side for experiment 2? Use this value to find the percent error in the net torque.
Explanation / Answer
A) for experiment 1
the net torque on left side is Rl X Fl = 0.1157 * 0.1381 = 0.0159 T
the net torque on right side is Rr X F = 0.0657 * 0.241 = 0.01583
B )
theoritical lever arm = torque by one/ force by 2 = 0.0159 T/0.0657 =0.242 or 24.2 cm
% error = (error/correct)*100 = (0.1/24.2)*100 = 0.41 %
C) the net torque on left side is Rl X Fl + Rl2+Fl2 = 0.0657 * 0.199 + 0.1157 * 0.099 = 0.0242 T
the net torque on right side is Rr X Fr = 0.1157 * 0.2128 = 0.024 T
D) theoritical lever arm = torque by one/ force by 2 = 0.0242 T/0.1157 =0.209 or 20.9 cm
% error = (error/correct)*100 = (0.38/20.9)*100 = 1.81 %
