Magnitude of acceleration of a speck of clay

AI Thread Summary
The discussion centers on calculating the magnitude of acceleration for a speck of clay on a potter's wheel rotating at 48 revolutions per minute with a diameter of 34 cm. The initial attempts involved using the formulas V=2(3.14)r/T and ar = v²/r, but the user encountered errors due to unit conversion issues. After realizing the mistake of not converting centimeters to meters, the user successfully arrived at the correct answer. The conversation emphasizes the importance of unit consistency in physics calculations. Accurate calculations are crucial for determining the acceleration in rotational motion.
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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