Consider an point particle moving on a frictionless semicircular hill (curve). The particle's initial kinetic energy is equal to the potential energy on the top of the hill, i.e it has the necessary energy to climb the hill.
Will it reach the top of the hill in infinite or finite time?
In my proof it needs infinite time and this is quite non-intuitive because, though the particle has the necessary energy to climb the hill, it needs an infinite time. Also if we reverse the time, when the particle is in equilibrium on the top of the hill, it will never go down, therefore this process is time irreversible.
Answer
EDIT: OK, I misinterpreted ANKU's question and had a bowl shape in mind whereas he had an upturned bowl in mind. This changes the energy equation to
$$ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R (\sin \theta) \; .$$
I measure the angle from the horizontal here. After similar manipulations as below, I get
$$ T = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\sin\theta}}$$
which is indeed a divergent integral. However, this does not mean that the motion is irreversible. The reverse motion starts from the top of the hill, but the top of the hill is an equilibrium point, albeit an unstable one, which implies that it takes an infinite amount of time to roll down the hill.
From conservation of energy, you can write down the following formula:
$$ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R (1-\cos \theta) \; .$$
The left hand side represents the potential energy at the top of your semi-circular hill, the right hand side the total energy at any point of the trajectory. (Angle measured from the vertical.)
Rearranging, you can write this as
$$ \frac{2 g \cos \theta}{R} = \dot{\theta}^2$$
or after some additional work and intergrating
$$ T = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{\cos\theta}}$$
A quick check with wolframalpha gives a finite number for the right integral. Therefore the time it takes for the ball to roll up the hill is finite.
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