THE PRACTICE STUDIO
Go slowly. Ask why. Make it yours.
PRACTICE / Physical pendulum
A swinging rod is not a point bob
A uniform thin rod of length 0.60 m swings through small angles about a horizontal axis through one end. Its moment of inertia about this pivot is ML²/3. Take g = 10 m/s².
Find the period.
Organize the givens & assumptions
- L = 0.60 m
- I = ML²/3
- Center of mass is L/2 from pivot
- g = 10 m/s²
Small angles in radians, fixed frictionless pivot, no damping and uniform gravity.
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Problem: A swinging rod is not a point bob
A uniform thin rod of length 0.60 m swings through small angles about a horizontal axis through one end. Its moment of inertia about this pivot is ML²/3. Take g = 10 m/s².
Question: Find the period.
Givens:
- L = 0.60 m
- I = ML²/3
- Center of mass is L/2 from pivot
- g = 10 m/s²
Assumptions: Small angles in radians, fixed frictionless pivot, no damping and uniform gravity.
Choices (LaTeX):
A. 0.20\pi\ \mathrm s
B. 0.40\pi\ \mathrm s
C. 2\pi\sqrt{0.06}\ \mathrm s
D. 2\pi\sqrt{0.03}\ \mathrm s
Ask for my approach first. Help me distinguish a physical misconception from an algebra error without revealing which choice is correct.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.Start with the smallest hint.
Work through it with me.
One decision at a time. Pause, predict, then unpack the algebra.
Use inertia about the pivot and weight acting at the center of mass.
Write the small-angle torque equation
Weight provides restoring torque about the pivot.
Why is this step valid?
The lever arm to the center of mass is L/2, while the inertia is about the end.
Identify angular frequency
Divide by the end-axis inertia.
Why is this step valid?
The total mass cancels; the small-angle linearization produces harmonic motion.
Convert to period
One cycle spans 2π radians of phase.
Why is this step valid?
Radians are dimensionless, so the period is in seconds.
The equivalent simple-pendulum length is 2L/3 = 0.40 m.
TRY THE IDEA AGAIN
A rod with the same shape but four times the length has a period…
YOUR LEARNING NOTE
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