Sphering inside a sphere!

Classical Mechanics Level 3

A particle is projected, along inside the inside of a smooth fixed sphere, from its lowest point, with a velocity of $20\text{ m/s}$ .

The particle, eventually, leaves the sphere's surface and afterwards passes through point $P$ which is vertically over the point of projection.Find the distance between $P$ and the point of projection.

Take radius of the sphere to be 10 m and value of $g=10 \text{ m/s}^2$ .

The answer is 15.625.

1 solution

Prakhar Bindal
Dec 15, 2016

Its a simple but nice problem . i had posted this kind of problem (in which the particle passes through the point of projection after leaving contact last year.

Well the problem is quite procedural.

Assume that at any angle x (the angle which line joining particle and centre makes with horizontal) the particle leaves contact .

Use Work energy theorem from bottom to this point to get the velocity at that point (initial velocity is sqrt(4gR))

You will get v^2 = 2gR(1-sinx)

Now as normal reaction will tend to zero as particle looses contact hence in radial direction gravity will be providing the centripetal force.

you get v^2 = gRsinx

Equate to get sinx = 2/3

Now use Equation of trajectory of the parabola to get the required result

Simple standard approach (+1)

Harsh Shrivastava - 4 years, 6 months ago

Sphering inside a sphere!

The answer is 15.625.

1 solution

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