Question

A bicycle wheel with diameter 16 inches rides over a screw in the street. the screw is on level ground before it punctures the bike’s tire. after the bike has moved forward another 56π inches, how high above the ground is the screw? round to the nearest tenth of an inch.

Answer 1

The height of the screw = 16 inches.

How do you determine the number of revolutions in a circle?

The total distance covered in one revolution will be equal to the perimeter of the wheel. Finally, to find the total number of revolutions, divide the total distance by distance covered in one revolution.

Given that,

Diameter of a bicycle = 16 inches

The distance the bike moves (forward) after the screw punctures the tire = 56π inches

We note that the circumference of the bicycle = π·D = π × 16 = 16π inches

Therefore,

$\frac{56\pi }{16\pi }$ = 3.5

Showing that the bicycle moves three and half complete turns (revolution) where after each complete turn, the screw starts from the bottom of the tire.

The height, h of the screw in the final half turn is given by the relation;

h = A cos(Bx - C) + D

Where

A = Amplitude of the motion = Diameter/2 = 16/2 = 8

P = The period of the motion 2π/B

Bx = The angle described by the motion = Half of one revolution = π = 180°

C = Phase shift = π

D = The mid line = Diameter/2 = 8 inches

Therefore;

h = 8×cos(π - π) + 8 = 16 inches

Hence, After the bike moves forward another 56π inches the height of the screw = 16 inches.

To learn more about number of revolution from the given link:

brainly.com/question/17266654

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