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

Two point mass 50 g and 100 g are attached to the ends of a rod of length 2 m and of negligible mass. The rod which was initially at rest is set into rotation about an axis perpendicular to the length of the rod. Find the position of axis of rotation from 50 g about which the work required to set the rod into rotation with an uniform angular velocity of 2 rad/s, is minimum.


A
23 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
43 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
13 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 43 m
Let the position of the axis of rotation from 50 g be x. About this axis of rotation, moment of inertia of system remains constant.
From work-energy theorem,
W=12Iw2


Here, moment of inertia of the system is
I=m1x2+m2(lx)2
=0.05(x)2+0.1(2x)2
=0.05x2+0.1[22+x22(2)x)]
I=0.15x2+0.40.4x

Now, for the work to be minimum, moment of inertia should be minimum
For I to be minimum, dIdx=0
ddx(0.15x2+0.40.4x)=0
0.3x0.4=0
x=43 m

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon