Question

a small spherical shell is slightly disturbed from the top of a large fixed hemisphere .if there is no slipping,find the angle with the vertical at which shell loses contact with the hemisphere . options are... a.cos ttheta =2/3 b. cos theta=3/5 c. cos theta =5/9 d. cos theta = 6/11​

Answer 1

Answer:

A small body of mass m slides down from the top of a hemisphere of radius r. There is no friction between the surface of the block and the hemisphere. The height at which the body loses contact with the surface of the sphere is?

This is how I understood the problem:

First of all, the mass does not lose contact with the surface of the sphere at all, considering it is undergoing centripetal acceleration, where the centripetal force is provided by the component of the mass's weight towards the centre of the hemisphere (mgcosθ, where θ is the angle between the vector of the mass's weight and its component acting towards the centre).

Therefore, since only mgcosθ is responsible for the centripetal force, I can form a relationship like this:

mgcosθ = mv2r

v = rgcosθ−−−−−√

Taking 'h' as the height of the mass from the base of the hemisphere.

cosθ = hr

Then the velocity of the mass becomes:

v = gh−−√

The component of the mass's weight along the centre disappears only when θ becomes 90 degrees. At this point, it leaves the surface of the hemisphere.

Now, the energy of the mass at the topmost point is:

P.E = mgr

As the body slides over the hemisphere's surface, it has a tangential velocity given by the expression I had just previously derived. So by the conservation of total mechanical energy of the body, its energy at any other point on the hemisphere is:

T.E = mgh+12mv2

P.E = T.E

gr = gh+v22

gr = gh+gh2

h = 23r