wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A wire of length 16 m and mass 8 kg is bent in the form of a rectangle ABCD with ABBC=2. The moment of inertia of this wire frame about side BC is


A
88.49 kg-m2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
108.5 kg-m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
50.5 kg-m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
38 kg-m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 88.49 kg-m2
Given,
Length of the wire (l)=16 m
Mass of the wire (m)=8 kg
and, ABBC=2
Perimeter of rectangle =2(AB+BC)
AB+BC=8
Since, AB=2BC, we get BC=83 m & AB=163 m
AD=BC=83 m and AB=CD=163 m

As the wire is uniform, linear mass density λ=ml=816=12 kg/m
Using this,
mAB=λ(AB)=83 kg;mBC=43 kg
By symmetry, mAB=mCD=83 kg and mBC=mAD=43 kg

Moment of inertia of rectangle ABCD about BC
I=IAB+IBC+ICD+IDA
=2(IAB)+IDA
(IAB=IBC;IBC=0)

IAB=mABAB23=[83(163)2.13]=204881
IDA=mDAAB2=43×(163)2=102427

I=2×204881+102427
=716881=88.49 kg-m2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon