Key relationship
xCM = (m₁x₁ + m₂x₂ + Mrod(x₁+x₂)/2)/(m₁+m₂+Mrod).
Physics · Class 11
Find the balance point
The centre of mass weights each position by its mass. A uniform rod contributes its own midpoint. Move either particle or change its mass to move the balance point toward the heavier side.
Open the interactive experiment →Find the balance point
நிறை மையம் ஒவ்வொரு நிறை மற்றும் அதன் நிலையைப் பொறுத்தது. அதிக நிறையுள்ள பக்கத்தை நோக்கி நிறை மையம் நகரும்.
Key notes, relationships and model limits
Remember
The centre of mass weights each position by its mass. A uniform rod contributes its own midpoint. Move either particle or change its mass to move the balance point toward the heavier side.
Common trap
The geometric midpoint is the centre of mass only for a symmetric mass distribution.
Model limits
Particles plus a uniform rod; coordinates share one axis. No external force is needed to define a centre of mass.
What you will learn
Locate the mass-weighted balance point, including an extended uniform body.
Try the experiment
Try both scenarios, change one control, and predict the result before checking the graph and readouts.
Think of it this way
Change one input while keeping the others fixed, as you would in a controlled experiment.
Worked example
2 kg at −2 m and 4 kg at 2 m give xCM = (−4+8)/6 = 0.667 m. A 6 kg rod adds its mass at 0 m and makes xCM = 0.333 m.
Why this matters
The model connects find the balance point with an observable reading. Check the equation and assumptions before applying it.
Before you start
SI units · algebra · graph reading
Check your understanding
Where is the centre of mass of 2 kg at −2 m and 4 kg at 2 m?
- At the geometric midpoint, 0 m
- At 0.667 m
- At 1.333 m
Read the explanation
(2×−2 + 4×2)/6 = 2/3 m.
Course connection: Class XI Chapter 6. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.