Doubts regarding block sliding on hemisphere

Question

I have come across the classic problem of a body sliding on a frictionless hemisphere from rest. The problem is to find the height at which the body "falls off" from the hemisphere.

I have seen answers in this site and everybody seems to say that particle looses contact when Normal reaction, $N=0$. I don't understand why.

Since centripetal force is given by:

$F_c = mgcos\theta- N$ , (Where $\theta$ is angle from vertical)

And Conservation of mechanical energy results in :

$F_c= 2mg(1-cos\theta)$

as $\theta$ increases, clearly $F_c$ increases.

So if centripetal force increases, why does the body fall off? Also Normal reaction is pointing outward, so if it is decreasing doesn't it support the circular motion?( Since circular motion is supported by inward forces) So if $N=0$ , why does the body stop circular motion, as if $F_c=0$?

Answer 1

Lets change the problem a little. Suppose the body was tied to the hemisphere like a rollercoaster is tied to a track. An suppose we slide the body around the hemisphere at a slow constant speed. That is, we also have a force tangent to the hemisphere holding the body back.

When the body starts out at the top, it is sitting on the hemisphere. It doesn't fall through because the hemisphere is made of some rigid material that exerts an upward force just strong enough to keep the body from falling through. Since the body moves in a circle at constant speed, the total force is centripetal. $F = mv^2/r$.

As the body slides below the equator, the body doesn't fall off because the track holds it. The total force is still centripetal. But now the normal force is toward the center. If it wasn't for the inward normal force, the body would fall off.

If you start from the top at a faster speed, the body will stay on the hemisphere if gravity larger than the centripetal force needed to move along the hemisphere. The outward normal force will oppose gravity and reduce the total force to exactly what is needed to follow the hemisphere.

As the body slides down, the component of gravity toward the center gets smaller. There will be a point where that component isn't bigger than the centripetal force needed to follow the hemisphere. At that point, the body will follow a straighter curve than the hemisphere, and fly off.

Answer 2

The two equations you've written down tell you two different things. The first one, $$ F_c = mg\cos\theta - N $$ is the amount of centripetal force that is actually being applied to the block. The second, $$ F_c= 2mg(1-\cos\theta) $$ tells you the amount of centripetal force that is necessary to keep the puck moving on a circular path.

In other words, if the puck stays on the circular path, then the centripetal force must increase with $\theta$. But $N > 0$ and $\cos \theta$ is decreasing, which means the actual amount of radial force applied will eventually decrease below the amount of force required to keep it on the path. Thus, there will be some value of $\theta$ where the puck leaves the surface.

Answer 3

Centripetal force is not a real force, it is a name given to real forces in inertial frame(no pseudo forces).

It is actually the gravitational force (part of gravitational force $ F_g - N$) that acts as centripetal force.

We need this "centripetal force" to keep the body moving along circular path.

For a fixed velocity, If this force is inadequate the object must revolve around a larger radius till the force balances with $ mv^2/R $ why? Because the change in velocity needed to maintain this circular path in dt interval is $ v*d\theta $ and $ a= v d\theta/dt = v\omega = v^2/R= F$.

Thats why when you loosen the grip while revolving a stone in horizontal circle(UCM) the object moves outwards, and if the force is reduced to zero the object flings off tangentially ($R=infinity$)

Now coming to the problem here $F_g$ is more so $ N $ must balance but as the velocity increases we need more centripetal force, therefore $N$ must decrease (otherwise the radius must increase ie less contact ie less $N$).

Answer 4

There are many conditions you can apply to solve problems, most check when something is biggest, smallest, or equal to zero. The equation you wrote assumes that we are staying on the sphere, so let's solve it in another way and contradict this assumption. Writing down the x component of the velocity we get: $$v_x=\sqrt{2gR(1-cos\theta)}cos\theta$$ Now, note that $v_x$ can't get smaller with time (what force would push it towards $-\hat{x}$?). The maximum velocity will be found from the derivative with respect to $\theta$: $$v_x'=\frac{\sqrt{gr}(3cos(\theta)-2)sin(\theta)}{\sqrt{2(1-cos(\theta)}}=0\rightarrow\theta=arccos(\frac{2}{3})$$ Which gives the same answer (The answers above are great, just another way to understand this problem)