# Answer to Question #122628 in Mechanics | Relativity for Shumail

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
1
2020-06-17T09:24:40-0400 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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