Consider a bicycle where the large central gear has a radius of 4 inches and the small back gear has a radius of 2 inches. how many inches would a point on the outer edge of the large gear travel in a 150º rotation? what degree of rotation will the small gear undergo when a point on its outer edge travels the same linear distance determined in part a? if the center of the small gear is rigidly attached to the center of a bicycle wheel with a radius of 10 inches, how many inches will the bicycle travel with a single rotation of the large gear? how does the distance you determined in part c change if the radius of the small gear is 1.5 inches and the radius of the large gear is 4.5 inches?

These are four questions and four answers:

Data:

large gear, R = 4 in
small gear, r = 2 in

Question 1.

Part A: How many inches would a point on the outer edge of the large gear travel in a 150º rotation?

Solution:

1) Circumferece of tthe large gear: 2πR = 2π(4in) = 8π in ≈ 25.13 in

2) 8π inches correspond to a full rotaion, i.e. 360°, so make a proportion to find the distance for 150° rotation:

  8π          x
= => x = (150°) * 8π / (360°) =  10.47 in
 360°      150°

approximately 10.5 in.

Question 2:

Part B. What degree of rotation will the small gear undergo when a point on its outer edge travels the same linear distance determined in part A?

Solution:

1) A point on the outer edge of the small gear will translate (travel) the same linear distance determined in the part A.

2) Each full rotaion (360°) will travel 2π * 2 in = 4π in ≈ 12.57 in

3) So, make this proportion:

 10.47 in         12.57
= 
     x                 360°

And solve for x: x = 10.47 in * 360° / 12.57 in = 299.86°

practically 300°.

Question 3.

Part C. If the center of the small gear is rigidly attached to the center of a bicycle wheel with a radius of 10 inches, how many inches will the bicycle travel with a single rotation of the large gear?

Solution:

1) A full rotation of the small gear will make a full rotationof the bycicle wheel.

2) A full rotaion of the large gear will result in this number of rotations of the small gear:

1 rotation of large gear * 2π (4 in) = x * 2π (2 in)

=> x = (4 in / 2 in)  = 2 rotaions

3) Each rotation of the small gear will lead to one full rotation of the wheel.

4) A full rotation of the wheel will lead to a translation equal to the circumference of the wheel, which is:

2π(radius) = 2π(10 in) = 62.83 in

Then, 2 rotation will be: 2 * 62.83 in = 125.66 in

about 125.66 in.

Question 4.

Part D) How does the distance you determined in part C change if the radius of the small gear is 1.5 inches and the radius of the large gear is 4.5 inches?

Solution:

1) one rotaion of the large gear => 2π (4.5 in) travel of a point on the outer edge

2) 2π(4.5 in) travel on the outer edge will translate into

2π(4.5 in) / 2π(1.5 in) = 4.5 / 1.5 = 3 rotations of the small gear

3) 3 rotations of the small gear = 3 rotations of the bycicle wheel

4) 3 rotations of the wheel => 3 * 2π(10 in) = 60 π ≈ 188.5 in

188.5 in

B

Step-by-step explanation:

As T is the midpoint of SU, ST=TU. 9x=5x+28, 4x=28, x=7. ST=TU=63 and SU=126