JEE Class main Answered
How to calculate moment of inertia in these type of problems explain please
Asked by nipunverma59 | 05 Apr, 2019, 01:10: PM
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 +Md2 .....(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
Answered by Thiyagarajan K | 05 Apr, 2019, 04:09: PM
Application Videos
JEE main - Physics
Asked by srkgb8018 | 15 May, 2024, 08:12: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by sohan123asiwal | 12 May, 2024, 05:52: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by pranav10022007 | 12 May, 2024, 10:05: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by anupriya23102006 | 09 May, 2024, 04:09: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by ashwinskrishna2006 | 07 May, 2024, 05:33: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by srkgb8018 | 05 May, 2024, 10:12: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by sandhyapallapu22 | 03 May, 2024, 04:32: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by gamingbadboy085 | 01 May, 2024, 06:28: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by bhargavreddynellaballi | 30 Apr, 2024, 08:18: AM
ANSWERED BY EXPERT