Question

A floor polisher has a rotating disk that has a 19-cm radius. The disk rotates at a constant angular velocity of 1.5 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 41 s, in order to buff an especially scuffed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Answer 1

The distance in meters of the disk move during 41s is

x= 73.41m

Step-by-step explanation:

The velocity angular is 1.5 rev/s

`1.5(rev)/(s) (41s)/(1)=61.5 rev`

The revolutions here are the number of rounds so to determinate how far in lineal distance

`x_(d)=N*\pi *r\\r=19cm (1m)/(100cm)=0.19m\\N=61.5\\x_(d)=61.5*2\pi *0.19m\\x_(d)=73.41m`

The distance in meters is x=73.41m