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 ($mgcos\theta$, where $\theta$ is the angle between the vector of the mass's weight and its component acting towards the centre).
Therefore, since only $mgcos\theta$ is responsible for the centripetal force, I can form a relationship like this:
$mgcos\theta\ =\ \large \frac{mv^2}{r}$
$ v\ =\ \sqrt{rgcos\theta}$
Taking 'h' as the height of the mass from the base of the hemisphere.
$cos\theta\ =\ \large \frac{h}{r}$
Then the velocity of the mass becomes:
$v\ =\ \sqrt{gh}$
The component of the mass's weight along the centre disappears only when $\theta$ 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:
$T.E\ =\ mgh + \frac{1}{2}mv^2$
$P.E\ =\ T.E$
$gr\ =\ gh + \frac{v^2}{2}$
$gr\ =\ gh + \frac{gh}{2}$
$h\ =\ \frac{2}{3}r$
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?
Best Answer
From what I remember $h=\frac{2r}{3}$ is indeed the corect answer.
This the incorrect part of your reasoning.
From what I understand your saying that at $θ = 90 degrees$ the radial component of the weight is zero therefore the blocks falls off. A non-circular analogy to this would be saying that that gravity always acts downward therefore it is imposible to throw any object, which is of course incorrect. After the block falls of it's weight is still "pulling" the block in the radial direction, however the tangental velocity component is too big i.e. $mgcosθ < \frac{mv^2}{r}$
For completeness sake it is worth noting that the statement $mgcosθ = \frac{mv^2}{r}$ isn't true while the block is still on the hemisphere, as you didn't take into account the contact/normal force between the block and the hemisphere. However not including it doesn't matter as the normal force is zero at the moment the blocks leaves the surface. Similarly $cosθ = \frac{h}{r}$ will only hold when the block is still no the hemisphere