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A bug crawls around on a horizontal turntable rotating with constant angular speed ω. The massof the bug is m and the coefficient of friction of the bug with the surface of the turntable is µ.Recall that Fstatic friction ≤ |µN | where N = mg is the normal force. The onset of slippage occurswhen Fstatic friction = |µN |. Gravity points downward. Ignore non-inertial effects due to Earth’srotation.a) The bug crawls with constant speed vr relative to the turntable in a radial path. How farfrom the center of the turntable can the bug crawl before starting to slip (in terms of ω, µ,and g)?b) The bug crawls with constant speed vr relative to the turntable in a circular path of radius bon the turntable. The circular path is concentric with the center of the turntable. For whatvalue of vr (in terms of ω, b, µ and g) will the bug start to slip if it crawlsi) In the direction of rotation?ii) Opposite of the direction of rotation?

Question

A bug crawls around on a horizontal turntable rotating with constant angular speed ω. The massof the bug is m and the coefficient of friction of the bug with the surface of the turntable is µ.Recall that Fstatic friction ≤ |µN | where N = mg is the normal force. The onset of slippage occurswhen Fstatic friction = |µN |. Gravity points downward. Ignore non-inertial effects due to Earth’srotation.a) The bug crawls with constant speed vr relative to the turntable in a radial path. How farfrom the center of the turntable can the bug crawl before starting to slip (in terms of ω, µ,and g)?b) The bug crawls with constant speed vr relative to the turntable in a circular path of radius bon the turntable. The circular path is concentric with the center of the turntable. For whatvalue of vr (in terms of ω, b, µ and g) will the bug start to slip if it crawlsi) In the direction of rotation?ii) Opposite of the direction of rotation?

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Solution

a) The bug will start to slip when the centrifugal force equals the maximum static friction. The centrifugal force is given by F = mω²r, where r is the distance from the center of the turntable. The maximum static friction is given by F = µN = µmg. Setting these two equal gives mω²r = µmg. Solving for r gives r = µg/ω².

b) i) When the bug is moving in the direction of rotation, it will start to slip when the sum of the centrifugal force and the force due to the bug's motion equals the maximum static friction. The force due to the bug's motion is given by F = mv²/r, where v is the speed of the bug relative to the turntable. The centrifugal force is given by F = mω²r. The maximum static friction is given by F = µN = µmg. Setting these equal gives mv²/r + mω²r = µmg. Solving for v gives v = sqrt((µg - mω²r)r/m).

ii) When the bug is moving opposite the direction of rotation, it will start to slip when the difference between the centrifugal force and the force due to the bug's motion equals the maximum static friction. The force due to the bug's motion is given by F = mv²/r, where v is the speed of the bug relative to the turntable. The centrifugal force is given by F = mω²r. The maximum static friction is given by F = µN = µmg. Setting these equal gives mω²r - mv²/r = µmg. Solving for v gives v = sqrt((µg + mω²r)r/m).

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