When the pot is on the point of slipping

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The discussion focuses on calculating the rotation speed of a pottery platform when a pot is on the verge of slipping. The pot's distance from the center is 9 cm, and the coefficient of static friction is given as µs = 0.3. The key equations involved are the friction equation, which connects friction force to the friction coefficient and the pot's weight, and the centripetal acceleration equation for uniform circular motion. The calculated rotation speed when the pot is about to slip is 0.91 revolutions per second. Understanding and applying these equations is essential for solving the problem effectively.
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A student in a pottery class leaves a freshly baked clay pot to cool down on a rotating
platform. If the pot is positioned at 9cm away from the center, how fast does the platform
rotate (in units of revolutions per seconds) when the pot is on the point of slipping?
Assume that the coefficient of static friction between the pot and the platform is µs = 0.3.


answer -->0.91 rev/sec
i do not really understand this question or what is the right formula to use :frown::confused:
 
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jk2455 said:
A student in a pottery class leaves a freshly baked clay pot to cool down on a rotating
platform. If the pot is positioned at 9cm away from the center, how fast does the platform
rotate (in units of revolutions per seconds) when the pot is on the point of slipping?
Assume that the coefficient of static friction between the pot and the platform is µs = 0.3.


answer -->0.91 rev/sec
i do not really understand this question or what is the right formula to use :frown::confused:

The relevant equations are the friction equation (relates friction force to the friction coefficient and the normal force [weight] of an object), and the equation for centripital acceleration in uniform circular motion. Does that help? Write down those equations, and see if that gets you going on this problem.
 
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