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PhysicsRotationCalculation of M.I.Hard2 minPYQ_2023
PhysicsHardnumerical

ICM is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. IAB is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 23R from center, where R is the radius of the disc. The ratio of IAB and ICM is x:9. The value of x is ______.

Question diagram: I CM is moment of inertia of a circular disc about an axis (
Answer:
17.00
Solution:

We have,  ICM=12MR2 

Since AB is parallel to diameter and at a distance of 23R from it, using parallel axis theorem,

IAB=mR22+m2R32=1718mR2

Thus, the ratio is 

IABICM=1718MR212MR2=179

Thus, x=17

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.
2mℹ️ Source: PYQ_2023

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