a floor polisher has a rotating disk that has a 15cm radius the disk rotates at a constant angular velocity of 1.4rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in place for 45s, in order to beff an especially scuffed area of the floor. how far (in meters) does the spot on the outer edge of the disk move during this time
To find the distance the spot on the outer edge of the disk moves during 45 seconds, we need to calculate the circumference of the circular path followed by the spot on the rotating disk.
The circumference of a circle is given by the formula C = 2πr, where C is the circumference and r is the radius. In this case, the radius is 15 cm, which is equivalent to 0.15 meters.
C = 2π(0.15) = 0.3π meters
Now, we need to find how many revolutions the disk completes in 45 seconds. Since the angular velocity is given as 1.4 rev/s, we can multiply it by the time in seconds to get the number of revolutions:
Number of Revolutions = 1.4 rev/s * 45 s = 63 revolutions
To find the total distance traveled by the spot on the outer edge of the disk, we multiply the circumference by the number of revolutions:
Distance = 0.3π meters/rev * 63 revolutions = 18.84π meters
Considering that π is approximately equal to 3.14, we can evaluate the final result:
Distance ≈ 18.84 * 3.14 ≈ 59.24 meters
Therefore, the spot on the outer edge of the disk moves approximately 59.24 meters during the 45 seconds.