A floor polisher has a rotating disk that has a 17-cm radius. The disk rotates at a constant angular velocity of 1.3 revs/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 50 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?
The distance traveled by a spot on the outer edge of the disk can be calculated using the formula:
Distance = Radius x Circumference
The circumference of a circle can be calculated using the formula:
Circumference = 2πr
where r is the radius of the circle.
In this case, the radius of the disk is 17 cm. Therefore, the circumference of the disk is:
Circumference = 2π(17) = 34π cm
Since the disk rotates at a constant angular velocity of 1.3 revs/s, it completes 1.3 revolutions every second. Therefore, in 50 seconds, the disk completes:
1.3 revs/s x 50 s = 65 revolutions
Since the circumference of the disk is covered during each revolution, the distance traveled by a spot on the outer edge of the disk during the 50 seconds is:
Distance = Circumference x Number of Revolutions
Distance = 34π cm x 65 = 2210π cm
Therefore, the spot on the outer edge of the disk moves approximately 2210π cm during the 50 seconds.