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?

Question

Chad Bryant

Physics

16 May, 14:41

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?

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Answers (2)

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Carly Dunlap · 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.

Mcbride · 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