Circluar Motion Involving Rotations Per Minute

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SUMMARY

The discussion centers on calculating the magnitude of acceleration for a speck of clay on a potter's wheel rotating at 40 RPM with a diameter of 35 cm. The correct formula for linear velocity is v = 2πr/T, where T is the period of rotation. The user initially calculated the period as 0.67 seconds, which was challenged by another participant, suggesting it should be 1.5 seconds. The final acceleration is derived from the formula a = (v^2)/r, emphasizing the importance of accurate period calculation in circular motion problems.

PREREQUISITES
  • Understanding of circular motion principles
  • Familiarity with the formulas for linear velocity and centripetal acceleration
  • Basic knowledge of radians and revolutions
  • Ability to perform unit conversions (e.g., from RPM to seconds)
NEXT STEPS
  • Study the relationship between RPM and period in circular motion
  • Learn how to derive centripetal acceleration from linear velocity
  • Explore the effects of radius on acceleration in circular motion
  • Practice problems involving circular motion with varying diameters and speeds
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts related to rotational dynamics.

fishyfish
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Homework Statement



What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 40 rpm (revolutions per minute) if the wheel's diameter is 35 cm?

Homework Equations



v=2(pi)(r)/t a=(v^2)/r

The Attempt at a Solution



2(3.14)(.175)/.67 = 1.099

(1.099^2)/.175=a
 
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Welcome to PF.

Isn't your period 1.5 instead of .67?

You only get 40 revs per min so a rev should be longer than a sec no?
 

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