A coin is sliding down on a smooth hemi spherical surface of radius r .the height from the top from where it losses contact?
Answers
Answered by
14
A small body of mass mm slides down from the top of a hemisphere of radius rr. 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 (mgcosθmgcosθ, where θθ is the angle between the vector of the mass's weight and its component acting towards the centre).
Therefore, since only mgcosθmgcosθ is responsible for the centripetal force, I can form a relationship like this:
mgcosθ = mv2rmgcosθ = mv2r
v = rgcosθ−−−−−√v = rgcosθ
Taking 'h' as the height of the mass from the base of the hemisphere.
cosθ = hrcosθ = hr
Then the velocity of the mass becomes:
v = gh−−√v = gh
The component of the mass's weight along the centre disappears only when θθ becomes 9090degrees. At this point, it leaves the surface of the hemisphere.
Now, the energy of the mass at the topmost point is:
P.E = mgrP.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+12mv2T.E = mgh+12mv2
P.E = T.EP.E = T.E
gr = gh+v22gr = gh+v22
gr = gh+gh2gr = gh+gh2
h = 23rh = 23r
But this would mean that the body does indeed leave the surface of the hemisphere. It just doesn't add up. Can someone please explain if my approach and assumptions are valid and how I got to this completely contradictory answer?
PLZ mark me as a BRAINLIST
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 (mgcosθmgcosθ, where θθ is the angle between the vector of the mass's weight and its component acting towards the centre).
Therefore, since only mgcosθmgcosθ is responsible for the centripetal force, I can form a relationship like this:
mgcosθ = mv2rmgcosθ = mv2r
v = rgcosθ−−−−−√v = rgcosθ
Taking 'h' as the height of the mass from the base of the hemisphere.
cosθ = hrcosθ = hr
Then the velocity of the mass becomes:
v = gh−−√v = gh
The component of the mass's weight along the centre disappears only when θθ becomes 9090degrees. At this point, it leaves the surface of the hemisphere.
Now, the energy of the mass at the topmost point is:
P.E = mgrP.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+12mv2T.E = mgh+12mv2
P.E = T.EP.E = T.E
gr = gh+v22gr = gh+v22
gr = gh+gh2gr = gh+gh2
h = 23rh = 23r
But this would mean that the body does indeed leave the surface of the hemisphere. It just doesn't add up. Can someone please explain if my approach and assumptions are valid and how I got to this completely contradictory answer?
PLZ mark me as a BRAINLIST
Similar questions