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A glass of water is sitting a few centimeters from the center of a small circular spinning table that is turning with constant angular speed. What is providing the radial force necessary to hold the glass in place without slipping as the table rotates around?

User Nikolay R
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Final answer:

The friction between the glass and the spinning table provides the centripetal force needed to keep the glass in a circular path, as per Newton's third law.

Step-by-step explanation:

The question asks about the radial force that holds a glass in place on a spinning table. When an object is undergoing uniform circular motion, it requires a centripetal force to keep it moving in a circular path. On a frictionless table, this force could be provided by a cord tied to a pivot at the center, but in the case of a glass on a spinning table, the friction between the glass and the table is providing the necessary centripetal force. Without this friction, the glass would slide off due to inertia, which would carry it in a straight line tangent to the circular path.

Newton's third law explains the interaction between the glass and the table: as the table applies a centripetal force via friction to the glass, the glass applies an equal and opposite force on the table. This law also explains that, in a frictionless scenario e.g., an object tied to a cord and spun around, the tension in the string provides the necessary centripetal force. If the object were to move at a constant speed without the string, there would be no centripetal force, thus it would not follow a circular path.

User Fattastic
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