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

Short Answer

Expert verified

Themagnitude of the acceleration of a speck of clayis 3.89m/s2.

Step by step solution

01

Step 1. Understanding the velocity and acceleration of a speck of clay 

The speck of clay is rotating on the potter鈥檚 wheel. The speed of clay is dependent on the radius of the wheel and the time taken by the wheel.The acceleration of the clay varies with the velocity of the clay and the radius of the wheel.

02

Step 2. Identification of given data 

The given data can be listed below as,

  • The rotation of the wheel is, N=45rpm.
  • The diameter of the wheel is, D=35cm1m100cm=0.35m.
03

Step 3. Determination of the speed of the clay

The speed of the clay can be expressed as,

v=2rTv=2D21N

Here, r is the radius of the wheel, T is the time taken by the wheel.

Substitute the values in the above equation.

v=20.35m2145rpm60s1minv=0.825m/s

04

Step 4. Determination of the centripetal acceleration of a speck of clay

The centripetal acceleration can be expressed as,

ac=v2r

Here, v is the speed of the clay, r is the radius of the potter鈥檚 wheel.

Substitute the values in the above equation.

ac=0.825m/s20.35m23.89m/s2

Thus, the magnitude of the acceleration of a speck of clayis 3.89m/s2.

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