Question #122628

Close to the center of a campus is a tall silo topped with a hemispherical cap. The cap is

frictionless when wet. Someone has somehow balanced a pumpkin at the highest point. The

line from the center of curvature of the cap to the pumpkin makes an angle

frictionless when wet. Someone has somehow balanced a pumpkin at the highest point. The

line from the center of curvature of the cap to the pumpkin makes an angle

Expert's answer

Let initially pumpkin was at top then it slides to point P.

Applying energy conservation law,

Energy at top = Energy at point P.

Energy at top is only due to it's potential energy.

Energy at point P is due to it's kinetic energy and potential energy.

So,

"mgR = mgRcos\\theta + \\frac {mv^2}{2}"

"2gR(1-cos\\theta) = v^2" . . . . . . . . . . . (i)

Now, at point P, balancing forces so that it keep contact with hemisphere.

"mgcos\\theta = \\frac {mv^2}{R}"

"gRcos\\theta = v^2" . . . . . . . . . . . . (ii)

solving (i) and (ii)

"2gR(1-cos\\theta) = gRcos\\theta"

"2 - 2cos\\theta = cos\\theta"

"2 = 3cos\\theta\\implies cos\\theta = (\\frac{2}{3})"

"\\theta" is angle with vertical.

Numerically, "\\theta = 48.2^0"

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