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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?

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  • $\begingroup$ First of all, don't forget the factor of 1/2 when dealing with energy conservation .. Energy conservation implies,for your initial conditions, $r = \sqrt{2gh}$. Then, I urge you to reconsider your statement that the ball disconncets at $\theta=90$. Remember you want to keep the ball on an arc of radius $R$. As it slides down, it gains velocity. At some point, the radial component of the gravity is just not enough to provide the acceleration to keep the ball on an arc. $\endgroup$ – Mathusalem Aug 19 '13 at 19:13
  • $\begingroup$ @Mathusalem. Could you please elaborate by giving an answer to this question? I did not understand. $\endgroup$ – Ram Sidharth Aug 19 '13 at 19:27
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From what I remember $h=\frac{2r}{3}$ is indeed the corect answer.

This the incorrect part of your reasoning.

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.

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

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As the body slides around the hemisphere, it drops lower, from its original height of $r$ to a height of $r-r\cos\theta$, where $\theta$ starts at 90 degrees and drops to 0 degrees (It's analogous to latitude.)

So the loss of potential energy is equal to the kinetic energy:$$\frac12mv^2=mgr(1-\cos\theta)$$ So $$v^2=2gr(1-\cos\theta)$$

The inward centripetal force needed is indeed given by:$$F_C=\frac{mv^2}{r}$$so, substituting:$$F_C=\frac{2mgr(1-\cos\theta)}{r}=2mg(1-\cos\theta)$$ The inward centripetal force available due to the radial component of gravity is $mg\cos\theta$, minus the normal outward force of the surface acting on the block. When the block "needs" all the gravity component, and then more, it leaves the surface (unless the surface can "suck"!

Set the radial component, $mg\cos\theta$ (yes, the same $\theta$) equal to the force needed and solve...

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  • $\begingroup$ Thank you very much for your answer. It has cleared a lot of doubts. But I have some questions. We know that $mgcos\theta \ =\ R_h$. $R_h$ becomes $0$ when $cos\theta$ becomes $0$. This would imply that the mass has slid all the way down to the foot of the hemisphere. But the answer that I get eventually is $h\ =\ \frac{2}{3}r$ which would mean it leaves the surface much before. The answer is contradictory, or maybe I'm interpreting this wrong. Could you clarify? $\endgroup$ – Ram Sidharth Aug 19 '13 at 21:00
  • $\begingroup$ We do not know that $mgcos\theta \ =\ R_h$ We know how large the force of gravity is; nothing can change that. We know the radial component of that gravity force; simple trig gives that. Theory tells us how much of that radial force is needed to keep the block moving in a circle. If the radial component of gravity is large enough to keep the circular motion, then fine; the reaction force is just big enough to offset the excess radial gravity force. If the radial force is just a tiny bit too small to keep the circular motion circular, then ggodbye block... $\endgroup$ – DJohnM Aug 20 '13 at 1:13
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Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.

Let a constant force F- be applied on the body such that it makes an angle qwith the horizontal and body is displaced through a distance s

By resolving force F- into two components:

(i) F cosq in the direction of displacement of the body.

(ii) F sinq in the perpendicular direction of displacement of the body.

Since body is being displaced in the direction of F cosΘ, therefore work done by the force in displacing the body through a distance s is given by

w = (f cosΘ)s = fs cosΘ

or w = fs

Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.enter image description here

Source:http://www.transtutors.com/homework-help/civil-engineering/work-power-and-energy/

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protected by AccidentalFourierTransform Nov 2 '17 at 21:27

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