Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

How to calculate moment of inertia in these type of problems explain please
question image
Asked by nipunverma59 | 05 Apr, 2019, 13:10: PM
answered-by-expert Expert Answer
In case-1, all spheres are 25 cm from axis of rotation.
 
By parallel axis theorem, moment of inertia of sphere whose axis of rotation is parallel to axis of sphere = (2/5)MR2 +Md.....(1)
 
where M is mass of sphere, R is radius of sphere and d is distance between axis of rotation and axis of sphere.
 
Using eqn.(1), Moment of Inertia I1 of 4 solid spheres for case-1 = 4×[ (2/5)×1×102 ×10-4 + 1×252×10-4 ] = 0.266 kgm2 ............(2)
 
In case-2 two spheres are in axis of roattion and other two spheres are 25√2 cm away from axis of rotation.
 
Hence moment of inertia I2 for case-2  = 2×[ (2/5)×1×102×10-4 ] + 2×[ (2/5)×1×102×10-4 + 1×(25√2)2×10-4 ] = 0.266 kgm2 ..........(3)
 
from (2) and (3), we have  I1 = I2  or  begin mathsize 12px style I subscript 1 over I subscript 2 space equals space 1 end style
Answered by Thiyagarajan K | 05 Apr, 2019, 16:09: PM
JEE main - Physics
Asked by rambabunaidu4455 | 03 Oct, 2024, 16:03: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by ratchanavalli07 | 17 Sep, 2024, 07:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by adithireddy999 | 03 Sep, 2024, 09:35: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vaishalinirmal739 | 29 Aug, 2024, 18:07: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vradhysyam | 26 Aug, 2024, 17:17: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT