Question

A small block of mass m is placed at the bottom of fixed circular smooth surface of radius R as shown in the figure. If a velocity v = √(4gR) is given to the block then maximum height from the bottom of circular surface, where block will leave the contact with the surface is
(1)5R/2
(2)3R/2
(3)R
(4)R/2​

Answer 1

Answer:

From the equation of motion of the block:

R

mv

2

=N−mg sinθ

or, N=

R

mv

2

+mg sinθ

By energy conservation, we get:

mg R sinθ=

2

1

mv

2

or,

R

mv

2

=2mg sinθ

Using this, we get:

N=3mgsinθ

So, ratio=

RN

mv

2

=

3

2

(constant)

x=

3

2

and is independent of the angle θ

Explanation:

Hope this helps mate.