Circluar Motion Involving Rotations Per Minute

Thread starter fishyfish
Start date Jan 7, 2009
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Motion Per Rotations

AI Thread Summary

To calculate the acceleration of a speck of clay on a potter's wheel turning at 40 rpm with a diameter of 35 cm, the relevant equations for linear velocity and centripetal acceleration are used. The radius is derived from the diameter, leading to a calculation of velocity. There is a correction noted regarding the period of rotation, suggesting it should be 1.5 seconds instead of 0.67 seconds, as each revolution takes longer than a second given the 40 rpm rate. This discrepancy impacts the final calculation of acceleration. Accurate understanding of the period is crucial for determining the correct magnitude of acceleration.

Jan 7, 2009

fishyfish

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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Jan 7, 2009

LowlyPion

Homework Helper

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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