Question

A floor polisher has a rotating disk that has a 14-cm radius. The disk rotates at a constant angular velocity of 1.3 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 36 s, in order to buff an especially scuffed area of the floor. How far does a spot on the outer edge of the disk move during this time

Answer 1

Distance moved = 41.2m

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below

Answer 2

d = 41.17 m

The outer edge of the disk moves 41.17m during this time

Step-by-step explanation:

Given;

angular velocity w = 1.3 rev/s

Radius of rotating disc r = 14 cm = 0.14m

Time t = 36s

The linear velocity v can be derived using;

v = 2πwr

And the distance covered can be expressed as;

d = vt = 2πwrt

d = 2πwrt .......1

Where; w,t,r are angular velocity,time and radius respectively.

Substituting the given values into equation 1

d = 2π×1.3×0.14×36

d = 41.17 m

The outer edge of the disk moves 41.17m during this time