A small body slides down a smooth uneven surface from a height H which eventually emerges into a circular loop of radius R(
↑ \
H \ )--A
↓ ︶
Answers
Explanation:
To complete a smooth vertical track of radius R, the minimum height at which a particle starts, must be equal to
2
5
R (one can proved it from energy conservation). Thus in our problem body could not reach the upper most point of the vertical track of radius
2
R
. Let the particle A leave the track at some point O with speed v (figure shown below). Now from energy conservation for the body A in the field of gravity
mg[h−
2
h
(1+sinθ)]=
2
1
mv
2
or, v
2
=gh(1−sinθ) (1)
From Newton's second law for the particle at the point O, F
n
=mw
n
,
N+mgsinθ=
(
2
h
)
mv
2
But, at the point O the normal reaction N=0
So, v
2
=
2
gh
sinθ (2)
From (1) and (2), sinθ=
3
2
and v=
3
gh
After leaving the track at O, the particle A comes in air and further goes up and at maximum height of it's trajectory in air, it's velocity (say v
′
) becomes horizontal (figure shown below). Hence, the sought velocity of A at this point,
v
′
=vcos(90−θ)=vsinθ=
3
2
3
gh