Circular motion and static friction of earth

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SUMMARY

The discussion focuses on the dynamics of a man standing on a rotating platform on Earth, which rotates at a constant angular speed of 2 rad/s. Key calculations involve the centripetal acceleration, expressed as a = ω²r, where ω is the angular speed and r is the distance from the axis of rotation. The forces acting on the man include static friction providing centripetal force and the normal force balancing gravitational force. The analysis confirms that the man's weight remains constant regardless of his distance from the rotational axis, clarifying that the question posed is not a trick.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Knowledge of static friction and its role in preventing slipping
  • Familiarity with forces acting in two dimensions (x and y planes)
  • Basic grasp of angular velocity and centripetal acceleration
NEXT STEPS
  • Study the concept of centripetal force in rotating systems
  • Learn about the effects of angular velocity on perceived weight
  • Explore the implications of static friction in non-inertial reference frames
  • Investigate the growth direction of plants in rotating environments
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in the principles of circular motion and the effects of static friction in rotating systems.

brownie24
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Homework Statement


A platform on Earth rotates at constant angular speed 2 rad/s. A 60 kg man stands on the platform. Static friction between the platform and the man's feet prevent slipping.
(a) Make a diagram showing the forces he exerts on the platform standing a distance r from the axis of rotation. r = 0, 1, 2
(b) What direction does the man sense as "up"? Calculate the angle that "up" makes with the vertical using values of r as stated.
(c) If plants were planted on the platform, in what direction would they grow?

Homework Equations


a = omega^2 r

The Attempt at a Solution


(a) In the x plane, force of friction acts as force of centripetal acceleration = m omega^2 r. What I don't understand is what is the reactive force?
In the y plane, normal force opposes gravitational force = mg. If this is so, the man's weight should be constant, despite his distance from the rotational axis. Is (a) a trick question?
 
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brownie24 said:
(a) In the x plane, force of friction acts as force of centripetal acceleration = m omega^2 r. What I don't understand is what is the reactive force?
It doesn't matter what we call the reaction force merely that one exists. Since the table exerts a force on the man, the man must also exert a force on the table in the opposite direction (i.e. radially outward) and of the same magnitude. I am being intentionally vague regarding the 'name' of this reaction force to avoid a long discussion on the centrifugal force.
brownie24 said:
In the y plane, normal force opposes gravitational force = mg. If this is so, the man's weight should be constant, despite his distance from the rotational axis. Is (a) a trick question?
No trick, you are indeed correct.
 

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