1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A hollow vertical cylinder of radius R and height h has smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed vo tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will

A

h2πR

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
v02gh
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2πRh

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

v02πR(2hg)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D v02πR(√2hg) since the body has no initial velocity in the vertical direction. az=-g, vertical displacement z=-h. ∴z=at+12at2 ⇒−h=0+12(−g)t2 ⇒T=√2hg time taken to reach the bottom let,t be the time taken to complete one revolution. Then t=2πRv0 ∴ number of revolution=Tt=√2hg2πRv0=v02πR√2hg

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Relative
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program