An Open-Book 3-hour Test of Lab 2 of 2LC and Lab 3 of 40B
The Pendulum

You can only talk to your PARTNER(S) or the Teaching Assistant. If you have the leisure to show off your superior intellect, then I must follow the supervisor's instructions strictly: ask you all to do every thing required by the LAB MANUAL.

The NAME of EXAMINEE:

The NAME of EXAMINEE's partner:

The EXAMINEE's workstation:

Remark:

We use two unit systems in physics, SI unit system, or so-called MKS system, and CGS unit system.

“MKS” means “meter”, “kilogram”, and “second.” Hence, the SI unit of force is exactly kg-m/sec². We call it “Newton.”

“CGS” means “cm”, “gram”, and “second.” Therefore, the CGS unit of force would have to be g-cm/sec². People use to call it “dyne.”

Procedure - Ideal Pendulum - variation of T with m

For the ideal pendulum, T vs. m:

1, Complete all basic measurement.

QUESTION 1:
How might your measurement be affected by the mass of the string?

REQUIREMENT: Now, you must treat this string as a stiff rod and the metal ball as a point mass. And, please show all your work numerically!
• (0.5) What is I_total=I_{point mass}+I_rod, the total moment of inertia of this system, with respect to the pivot?

  ANSWER:

• (0.5) What is h, the distance from the pivot to the center of mass of this stiff-rod-and-point-mass system?

  ANSWER:

• (0.5) What will be the new time period formula if you treat this system as a physical pendulum, starting from T = 2p? Here m_total=m_{point mass} + m_rod is the total mass.

  ANSWER:

• (0.5) What is the percent difference of time period between the massless-string ideal pendulum and the stiff-rod physical pendulum?

  ANSWER:

Procedure - Ideal Pendulum - variation of T with q₀

QUESTION 2:
Suppose a clock has a swing amplitude of 0.1 radians. How much increase (or decrease) in amplitude would make the clock be off by one second in one day (86,400 seconds)?

REQUIREMENT: Now, to simplify the above question, according to TA manual, the answer of this question is q_f = 0.1009(radian) or 0.0991(radian).
However, please show all your work and prove the given answer!
• (0.5) Set the initial amplitude q_i = 0.1000 (radian) and define the unknown final amplitude as q_f. With T_i @ T₀[1+ q_i²/16] , please express DT=T_f-T_i, the difference of time period between q_f and q_i, DT, in terms of T₀, q_f and q_i!

  ANSWER:

• (0.5) From the above result, please express the final amplitude q_f in terms of DT/T₀ and q_i!

  ANSWER:

• (0.5) Read the phrase "off by one second in one day" carefully. Then write down the numerical value of DT/T₀.

  ANSWER:

• (0.5) Write down the numerical value of Dq = q_f - q_i. Prove the percent error in amplitude is 0.9%.

  ANSWER: