A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x - y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' will be :
Circular ring in x-y plane with axis yy' tangent to the ring Q20
Ring mapped on x-y axis showing yy' axis tangent to the boundary of the ring.

Solution & Explanation

Related Formula

M = m × L

L = 2π r r = (L)/(2π) Idiametric = (M r²)/(2) Itangent = ICM + M r²
Core Logic

Total mass of the ring M = m × L, where m is the linear mass density. The radius r of the ring is derived from the circumference: r = (L)/(2π).

The required axis yy' is a tangent to the ring in its own plane. By the parallel axis theorem, the moment of inertia about this tangential axis is:

Iyy' = Idiametric + M r² Iyy' = (M r²)/(2) + M r² = (3)/(2) M r²
Step 1: Substitute Length and Mass

Substitute M = mL and r = (L)/(2π) into the equation:

Iyy' = (3)/(2) (mL) ( (L)/(2π) )² Iyy' = (3)/(2) mL ( (L²)/(4π²) ) Iyy' = 3 ~mL³8π²
Pattern Recognition

Whenever finding tangential MoI, verify if the tangent is "in-plane" (I = (3)/(2)MR²) or "perpendicular to plane" (I = 2MR²). The figure strictly places the tangent in the x-y plane.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion

Reference Study Guides

Systems Of Particles And Rotational Motion

More Systems of Particles and Rotational Motion Previous-Year Questions

Q38 neet_2026_03_may_morning Kinematics of Rotational Motion about a Fixed Axis
The angular speed of a flywheel is increased from 600 ~rpm to 1200 ~rpm in 10 ~s. The number of revolutions completed by the flywheel during this time is :
  • A. 600
  • B. 900
  • C. 300
  • D. 150

Solution

Related Formula
θ = ( (ω₁ + ω₂)/(2) ) t ω = 2π f = 2π ( (N)/(60) )
Core Logic

We can solve this directly using average rotational speed since the acceleration is constant. Initial frequency N₁ = 600 ~rpm = (600)/(60) = 10 ~rev/s. Final frequency N₂ = 1200 ~rpm = (1200)/(60) = 20 ~rev/s. Time t = 10 ~s.

Step 1: Calculate Revolutions via Average Frequency

Since angular acceleration is constant, the average frequency (revolutions per second) is:

Navg = (N₁ + N₂)/(2) Navg = (10 + 20)/(2) = 15 ~rev/s

Total number of revolutions = Navg × t Total revolutions = 15 × 10 = 150 revolutions.

Pattern Recognition

Converting strictly into radians per second is technically correct but unnecessarily tedious. Staying in "revolutions per second (Hz)" allows directly multiplying average Hz by time to get total revolutions instantly.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion

More Systems of Particles and Rotational Motion Questions — neet_2026_03_may_morning

Practice all Systems of Particles and Rotational Motion previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)