Question

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 46 rpmrpm (revolutions per minute) if the wheel's diameter is 32 cmcm ?

Answer 1

Acceleration of a speck is 0.77 m/s²

Step-by-step explanation:

Given of the solution-

Diameter (We can represent as d) = 32 cm

radius, (We can represent as r )= 32/2 = 16 cm = 0.16 m

Angular acceleration,(We can represent as α ) = 46 rpm

`\alpha = 46 * (2\pi )/(60)`

Acceleration, a = ?

We know that the formula for the acceleration is

`a = r * \alpha`

`a = 0.16 * 46 * (2\pi )/(60)\\ \\a = 0.16 * 46 * (2 * 3.14)/(60)`

`a = 0.77m/s^2`

Therefore, acceleration of a speck is 0.77 m/s²