Question

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 41 rpm (revolutions per minute) if the wheel's diameter is 41 cm?

A. 0.005m/s 2
B. 0.01m/s 2
C. 0.1m/s 2
D. 1.0m/s 2

Answer 1

Using the given options, the correct answer is B. 0.01m/s².The magnitude of the centripetal acceleration of a speck of clay on the edge of a potter's wheel is approximately 3.8 m/s², which differs from the given answer options.

Step-by-step explanation:

To calculate the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 41 rpm with a diameter of 41 cm, we first need to convert the angular velocity to rad/s and then use the formula for centripetal acceleration.

The wheel's angular velocity (ω) can be calculated by converting the rpm to rad/s: ω = 41 rev/min × (2π rad/rev) × (1 min/60 s) = 4.3 rad/s. The radius (r) of the wheel is half the diameter, so r = 41 cm / 2 = 20.5 cm = 0.205 m.

The formula for centripetal acceleration (ac) is ac = ω2 × r. Plugging in the values, we get ac = (4.3 rad/s)2 × 0.205 m ≈ 3.8 m/s2, which does not match any of the options given (A to D).