From a Circular Disc of Radius r and Mass 9m – Complete Solution Guide
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| From a Circular Disc of Radius r and Mass 9m |
The phrase "from a circular disc of radius r and mass 9m" commonly appears in Physics and Engineering Mechanics problems involving the moment of inertia, center of mass, rotational dynamics, and removal of a smaller disc from a larger circular disc.
Many students encounter such questions in competitive examinations, including JEE Main, JEE Advanced, NEET, and various university entrance tests. These problems test conceptual understanding of rotational motion and mass distribution.
In this guide, we will understand how to approach questions based on a circular disc of radius r and mass 9m, the formulas involved, common shortcuts, and examination strategies.
Understanding the Circular Disc
A circular disc is a two-dimensional object having uniform thickness and mass distribution.
Given:
Radius = r
Since the mass is uniformly distributed, the surface mass density is constant throughout the disc.
Surface density:
σ = Total Mass / Area
Area of disc:
A = πr²
Therefore,
σ = 9m / πr²
This density concept becomes extremely useful when a smaller section is removed from the disc.
Moment of Inertia of a Circular Disc
One of the most frequently asked questions involving a circular disc of radius r and mass 9m relates to the moment of inertia.
Moment of inertia about the central axis:
I = ½MR²
Substituting:
M = 9m
R = r
Therefore,
I = ½ × 9m × r²
I = 9mr²/2
This is the moment of inertia about an axis passing through the center and perpendicular to the plane of the disc.
When a Smaller Disc is Removed
A common examination problem states:
"A smaller disc of radius r/3 is removed from a larger disc of radius r and mass 9m."
Step 1: Find Density
σ = 9m / πr²
Step 2: Area of Smaller Disc
Area = π(r/3)²
Area = πr²/9
Step 3: Mass of Smaller Disc
Mass = Density × Area
Mass = (9m/πr²) × (πr²/9)
Mass = m
Thus, a smaller disc of radius r/3 removed from the larger disc has mass m.
This result appears very frequently in objective examinations.
Center of Mass Concept
When material is removed from a symmetric body, the center of mass shifts.
To solve such problems:
Consider the removed portion as negative mass.
Use center of mass equations.
Calculate displacement from the original center.
Formula:
x = (Σmx)/(Σm)
The removed disc contributes negative mass.
This shortcut greatly simplifies calculations.
Parallel Axis Theorem
Another important concept used in circular disc problems is the parallel axis theorem.
Formula:
I = Icm + Md²
Where:
I = Moment of inertia about required axis
Icm = Moment of inertia about center of mass
M = Mass
d = Distance between axes
Students should remember this theorem because many questions involving removed discs require it.
Rotational Dynamics Applications
Circular disc problems frequently appear in rotational dynamics.
Important equations:
Torque:
τ = Iα
Angular momentum:
L = Iω
Rotational kinetic energy:
K = ½Iω²
When the moment of inertia changes due to removal of a smaller portion, all these quantities are affected.
Competitive Exam Perspective
Questions based on a circular disc of radius r and mass 9m are popular because they combine:
Geometry
Mass density
Rotational mechanics
Center of mass
Moment of inertia
Examiners often create variations involving:
Semicircular cuts
Circular holes
Eccentric removals
Composite bodies
Students who understand density-based calculations can solve such problems quickly.
Shortcut Method
Whenever a disc has uniform density:
Mass ∝ Area
Therefore:
Mass Ratio = Area Ratio
For example:
Large disc radius = r
Small disc radius = r/3
Area ratio:
π(r/3)² : πr²
= 1 : 9
Therefore:
Mass ratio = 1 : 9
If large mass = 9m
Small mass = m
This shortcut saves significant time in exams.
Common Mistakes
Ignoring Uniform Density
Students often use incorrect mass values without calculating density.
Wrong Radius Substitution
Always square the radius carefully.
Forgetting Negative Mass
For removed sections, use negative mass while calculating center of mass.
Incorrect Axis Selection
Read the question carefully before applying the moment of inertia formula.
Sample Numerical Example
Question:
A circular disc of radius r and mass 9m has a smaller disc of radius r/3 removed from it. Find the mass of the removed disc.
Solution:
Density:
σ = 9m/πr²
Area of smaller disc:
π(r/3)²
= πr²/9
Mass removed:
σ × area
= (9m/πr²) × (πr²/9)
= m
Answer:
Mass removed = m
Exam Preparation Tips
Memorize standard moment of inertia formulas.
Practice center of mass problems daily.
Learn area-to-mass conversion shortcuts.
Revise parallel axis theorem regularly.
Solve previous year JEE and NEET questions.
Frequently Asked Questions
What is the moment of inertia of a disc of radius r and mass 9m?
I = 9mr²/2
Why is density important?
Density helps determine the mass of removed portions.
How do we calculate the mass of a smaller removed disc?
Mass is proportional to area because density is uniform.
Which exams ask such questions?
JEE Main, JEE Advanced, NEET, BITSAT, and university engineering entrance examinations.
Problems involving a circular disc of radius r and mass 9m are among the most important topics in rotational mechanics. By understanding density, moment of inertia, center of mass, and rotational dynamics, students can solve even advanced variations quickly and accurately. Consistent practice and proper application of formulas are the keys to mastering these questions in competitive examinations.