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
Expert's answer
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=mgRcosθ+mv22mgR = mgRcos\theta + \frac {mv^2}{2}


2gR(1cosθ)=v22gR(1-cos\theta) = v^2 . . . . . . . . . . . (i)



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


mgcosθ=mv2Rmgcos\theta = \frac {mv^2}{R}


gRcosθ=v2gRcos\theta = v^2 . . . . . . . . . . . . (ii)



solving (i) and (ii)


2gR(1cosθ)=gRcosθ2gR(1-cos\theta) = gRcos\theta


22cosθ=cosθ2 - 2cos\theta = cos\theta


2=3cosθ    cosθ=(23)2 = 3cos\theta\implies cos\theta = (\frac{2}{3})


θ\theta is angle with vertical.


Numerically, θ=48.20\theta = 48.2^0



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