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

59.4 meters

Explanation:

The correct question statement is:

A floor 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 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t where t is time

or

θ=ωt

and we know

1 revolution = 2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec = 1.4*2π radians / sec = 1.4*3.14*2 * = 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ = 8.8 rad / sec * 4.5 sec

θ = 396 radians

We are given radius r = 15 cm = 15 * 0.01 m=0.15 m (because 1 m = 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r in

S=rθ

we have

S = 396 radians * 0.15 m=59.4 m