wiz-icon
MyQuestionIcon
MyQuestionIcon
15
You visited us 15 times! Enjoying our articles? Unlock Full Access!
Question

A light rod of length L can revolve in a vertical circle around point O. The rod carries two equal masses of mass m each such that one mass is connected at the end of the rod and the second mass is fixed at the middle of the rod. u is the velocity imparted to the end P to deflect the rod to the horizontal position. Again mass m in the middle of the rod is removed and mass at end P is doubled. Now v is the velocity imparted to end P to deflect it to the horizontal position. Then (uv)2 is

Open in App
Solution

In first case,
As velocity of mass attached to the middle of rod is u2
Using conservation of energy
12mu2+12m(u2)2=mgL+mgL2
u=(125gL)
In second case,
Using conservation of energy
12(2m)v2=2mgLv=(2gL)12
Now,
(uv)2=(125gL)[2gL]=[65]=1.2

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon