A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 45 s, in order to buff an especially scuff ed area of the fl oor. How far (in meters) does a spot on the outer edge of the disk move during this time?

59.4 m.

Explanation:

Using equations of angular motion,

S = r*θ

ω = θ/t

where

r = radius in meters

t = time in seconds.

θ = angular displacement in radians

ω = angular velocity in rad/s

Remember,

1 revolution = 2π radians.

ω = 1.4 rev/s

Converting rev/s to rad/s,

= 1.4 rev/s * 2π rads/rev

= 8.8 rad/s

t = 45 s

r = 15 cm

Converting cm to m,

= 15 cm x 10^-2 m/ 1 cm

= 0.15 m

ω*t = θ

θ = 8.8 * 4.5

θ = 396 radians

Remember,

S = r*θ

we have

S = 396 * 0.15

= 59.4 m

lol

explanation: