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.

The Length of the string = ___________________________, unit: (    )

The mass of the sample string = ___________________________, unit: (    )

The Length of the pendulum, L_pendulum = The Length of the string + The radius of the metal ball.

The following are Examples of Accuracy and Precision. Please refer to Lab 1 of Phys40A, or Lab 1 of Phys2A LAB.

Accuracy: (g_measured-g_true)/g_true ´ 100% = (9.732-9.80)/9.80´ 100% = -0.7%= -(7´ 10^-1)%

c_deg = 10^(-1+1) %= 1%

Precision: ±error/ g_measured ´ 100% = 0.086/9.732´ 100% = 0.88%= (8.8´ 10^-1)%

c_deg = 10^(-1+1) %= 1%

Ball	Mass unit: (    )	Diameter unit: (    )	Lpendulum unit: (    )	Theoretical Time Period T0, unit: (    )	Measured Time Period TMeasured ± SDOM unit: (    )		Precision %ERROR	Accuracy %ERROR
Aluminum				2p=		±	%	%
Iron						±	%	%

QUESTION 1-5-3:
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?
• (0.5) What is h, the distance from the pivot to the center of mass of this system, with ?
• (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.
• (0.5) What is the percent difference of time period between the massless-string ideal pendulum and the stiff-rod physical pendulum?

Procedure - Ideal Pendulum - variation of T with q₀

Preparing the Graph of "T versus q₀²"

we should have T @ T₀[1+ q₀²/16] = (T₀/16)q₀² + T₀ as Y = MX + B. Here T₀ = 2p.

For your reference, 5^o=8.72´ 10^-2(radian)

QUESTION 2-4-3:
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!
• (0.5) From the above result, please express the final amplitude q_f in terms of DT/T₀ and q_i!
• (0.5) Read the phrase "off by one second in one day" carefully. Then write down the numerical value of DT/T₀.
• (0.5) Write down the numerical value of Dq = q_f - q_i. Prove the percent error in amplitude is 0.9%.

PROCEDURE----Physical Pendulum

T = 2p = 2p