Four identical rods are joined end to end to form a square. The mass of each rod is M. The moment of inertia of the square about the central line yy′ as shown in figure is
A
Ml23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Ml24
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Ml26
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none of these
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D none of these From given, Moment of inertia about YY′ IYY′=IAB+IBC+ICD+IDA ....(1) here, road AB and CD are symmectric. so, moment of inertia about YY′ IYY′=Ml212 I(YY′forABandCD)=Ml212×2=Ml26
Moment of inertia for rods AD and BC dI=∫dm(l2)2 dI=l24∫dm IYY′=ML24 for rod AD and BC IYY′(ADandBC)=ML24×2=ML22
by using eq. (1) IYY′=IAB+IBC+ICD+IDA IYY′=Ml26+ML22=2ML23