Question

A coin is sliding down on a smooth hemi-spherical
surface of radius R. The height from the bottom.
where it looses contact with the surface is​

Answer 1

Answer:

Also, when the coin detaches from the surface the normal force on it will be zero. Let us look at the free body diagram. Therefore, we found that the height from the bottom, where it loses contact with the surface as23R.

Explanation:

Answer 2

Answer:

2R/3

Here, initial velocity of projection is zero.

Given, radius of hemisphere =R .

To find → value of h .

(h is height of Q from ground. )

It is clear, that h=Rcosθ

thus, the blocks covers (R−h)=R(1−cosθ) vertically

before leaving the contact of Q .

Let, the velocity attained by block at Q be v ,

then, v

2

=0

2

+2g(R−h)⇒v

2

=2gR(1−cosθ)

Now,

Now, A+B

→mgcosθ−N=

R

mv

2

(But N=0 at B , as contact

⇒gcosθ=

R

v

2

=2g(1cosθ)[∵v

2

=2gR(1cosθ)]

⇒cosθ=

3

2

Hence, h=Rcosθ=[

3

2R

]