Magnitude of acceleration of a speck of clay

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SUMMARY

The magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 48 revolutions per minute (RPM) with a diameter of 34 cm can be calculated using the formula for centripetal acceleration. The correct approach involves converting the diameter to meters, calculating the radius, and then applying the formulas V=2πr/T and a_r = v²/r. After correcting the unit conversion from centimeters to meters, the user successfully arrived at the correct answer.

PREREQUISITES
  • Understanding of centripetal acceleration
  • Familiarity with the formulas V=2πr/T and a_r = v²/r
  • Knowledge of unit conversion (cm to m)
  • Basic skills in algebra for solving equations
NEXT STEPS
  • Study the principles of circular motion and centripetal force
  • Learn about unit conversions in physics
  • Practice problems involving rotational dynamics
  • Explore the implications of angular velocity in real-world applications
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of rotating systems.

MannyB
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What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 48 (revolutions per minute) if the wheel's diameter is 34 ?

I tried using V=2(3.14)r/T then using ar =v2/r but i keep getting the wrong answer.
 
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MannyB said:
What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 48 (revolutions per minute) if the wheel's diameter is 34 ?

I tried using V=2(3.14)r/T then using ar =v2/r but i keep getting the wrong answer.
Hi, Manny, welcome to PF!. You seem to be on the right track, why don't you show your numbers and someone will correct your math. What about the units??
 
...i forgot to convert cm to m:mad: i have the right answer now.
 

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