Answer to Question #122628 in Mechanics | Relativity for Shumail

Question

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

Answer 1

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, $\theta = 48.2^0$