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A hemispherical bowl of radius R rotates about its vertical axis of symmetry. A small block rests inside the bowl where the radius to the block makes angle theta with the vertical, and rotates with the bowl without slipping. If the coefficient of friction between block and bowl is mu, find the range of angular speed omega for which the block does not slip.

1. sqrt[ g(sin(theta) - mu*cos(theta)) / (R*sin(theta)*(cos(theta) + mu*sin(theta))) ] <= omega <= sqrt[ g(sin(theta) + mu*cos(theta)) / (R*sin(theta)*(cos(theta) - mu*sin(theta))) ]
2. sqrt[ g(cos(theta) - mu*sin(theta)) / (R*(sin(theta) + mu*cos(theta))) ] <= omega <= sqrt[ g(cos(theta) + mu*sin(theta)) / (R*(sin(theta) - mu*cos(theta))) ]
3. omega <= sqrt[ g*tan(theta) / R ]
4. sqrt[ g*sin(theta) / (R*cos(theta)) ] <= omega <= sqrt[ g / (R*sin(theta)) ]

Correct answer: sqrt[ g(sin(theta) - mu*cos(theta)) / (R*sin(theta)*(cos(theta) + mu*sin(theta))) ] <= omega <= sqrt[ g(sin(theta) + mu*cos(theta)) / (R*sin(theta)*(cos(theta) - mu*sin(theta))) ]

Solution

Resolving the normal force and limiting friction horizontally (centripetal) and vertically (weight) and using r = R*sin(theta) gives the two limiting speeds; below the lower limit the block slides down and above the upper limit it slides up.

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