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

A particle of mass m attached to a string of length l fixed at a point such that it can perform circular motion in a vertical plane. The particle is imparted a velocity of 8gl at the lowest position by a sharp hit. Find the ratio of tension in string at the lowermost position to the tension in the string at the topmost position of the vertical circular path. Take g=10 m/s2.

A
3:1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9:5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
7:3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7:5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 3:1
Let tension at lowermost point be Tl and tension at topmost point be Tt

Applying equation of dynamics towards centre of circular path from the FBD at both positions:


At lowermost point,
Tlmg=mu2l
Tl=mg+8mgll=8mg+mg
Tl=9mg

Using energy conservation between lowermost and topmost position:
Kl+Ul=Kt+Ut
12mu2+0=12mv2t+2mgl
12m(8gl)=12mv2t+2mgl
v2t=4gl

At topmost point,
Tt+mg=mv2tl
Tt=4mgllmg=4mgmg=3mg

TlTt=9mg3mg=3
Tl:Tt=3:1

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