Question

What is the magnitude of the acceleration of a speck of clay on the edge of a potter’s wheel turning at 45 rpm (revolutions per minute) if the wheel’s diameter is 35 cm?

Answer 1

The magnitude of the acceleration of a speack of clay on the edge of potter's wheel turning at 45 rpm if the wheels diameter is 35cm ?

4.66 is your answer. 
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Answer 2

OK.  We can do this !  Fasten your seat belt.

The formula we need is:  A = V²/R
                           Centripetal acceleration = (speed)² / (radius) .

Also        V  =  D / T           Speed = (distance) / (time).

with everything in meters, kilograms, and seconds.

Circumference of the wheel = (π · diameter) = 0.35 π  meters.

Speed = (0.35π m) x (45 / minute) x (1 minute / 60 sec)

           = (0.35π · 45 / 60) m/sec

           =     0.2625π m/sec

Acceleration  =  (speed²) / (radius)

                    =  (0.2625π m/s)² / (0.175 m)

                    =  (0.06890625π² / 0.175)  m² / s²·m

                    =     3.89  m / s²