Question

A floor polisher has a rotating disk that has a 13-cm radius. The disk rotates at a constant angular velocity of 1.7 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 34 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

Answer 1

Solution:-

Given –

Radius of rotating disk – 13 cm

Angular velocity – 1.7rev/s

Time – 34 s

1.7rev/s*34s = 57.8 revolution

At the disk rotates one time, the spot on the outer edge moves the circumference of the circle.

C = 2*π*13

As the disk rotates 57.8 times the spot on the outer edge moves the 57.8times the circumference of the circle.

D = 57.8*3.14*2*13

D = 4718.79cm

D = 4719cm

D = 47.19m

The spot moves approximately 4719cm or 47.19m.

Solution:-

Given –

Radius of rotating disk – 13 cm

Angular velocity – 1.7rev/s

Time – 34 s

1.7rev/s*34s = 57.8 revolution

At the disk rotates one time, the spot on the outer edge moves the circumference of the circle.

C = 2*π*13

As the disk rotates 57.8 times the spot on the outer edge moves the 57.8times the circumference of the circle.

D = 57.8*3.14*2*13

D = 4718.79cm

D = 4719cm

D = 47.19m

The spot moves approximately 4719cm or 47.19m.