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A bicycle odometer (which counts revolutions and is calibrated to report distance traveled) is attached near the wheel axle and is calibrated for 27-inch wheels. What happens if you use it on a bicycle with 24-inch wheels?

Short Answer

Expert verified

The rotations of the smaller wheel of the bicycle will appear on the odometer as 9.43 inches more than the actual distance moved by the bicycle.

Step by step solution

01

Identification of given data

The given data can be listed below as:

  • The diameter of the wheels of one bicycle is\({d_1} = 27{\rm{ inches}}\).
  • The diameter of the wheels of another bicycle is\({d_2} = 24{\rm{ inches}}\).
02

Understanding the measurement of the distance on the odometer of the bicycle

The odometer measures the particular distance traveled by the bicycle. An odometer counts the revolutions of the bicycle wheel. One rotation of the wheel is equal to the value of the circumference of the bicycle. The circumference of the bicycle is equal to\(\left( \pi \right)\)times the diameter of the bicycle.

03

Determination of the distance moved by the bicycle

The distance traveled by the bicycle can be expressed as:

\({R_1} = \pi {d_1}\)

Here,\({d_1}\)is the diameter of the bicycle wheel.

Substitute the values in the above equation.

\(\begin{aligned}{c}{R_1} = \pi \times 27{\rm{ inches}}\\ = 84.82{\rm{ inches}}\end{aligned}\)

The distance traveled by another bicycle can be expressed as:

\({R_2} = \pi {d_2}\)

Here,\({d_2}\)is the diameter of another bicycle wheel.

Substitute the values in the above equation.

\(\begin{aligned}{c}{R_2} = \pi \times 24{\rm{ inches}}\\ = 75.39{\rm{ inches}}\end{aligned}\)

04

Determination of the extra distance moved by the 27 inches wheel 

The extra distance traveled by the 27 inches wheel of the bicycle can be given as:

\(R = {R_1} - {R_2}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}R = 84.82{\rm{ inches}} - 75.39{\rm{ inches}}\\ = 9.43{\rm{ inches}}\end{aligned}\)

Thus, the rotations of the smaller wheel of the bicycle will appear on the odometer as 9.43 inches more than the actual distance moved by the bicycle.

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Most popular questions from this chapter

The radius of the roll of paper shown in Fig. 8–67 is 7.6 cm and its moment of inertia is \(I = 3.3 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^2}\). A force of 3.5 N is exerted on the end of the roll for 1.3 s, but the paper does not tear so it begins to unroll. A constant friction torque of \(I = 0.11\;{\rm{m}} \cdot {\rm{N}}\) is exerted on the roll which gradually brings it to a stop. Assuming that the paper’s thickness is negligible, calculate (a) the length of paper that unrolls during the time that the force is applied (1.3 s) and (b) the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving

Let us treat a helicopter rotor blade as a long, thin rod, as shown in Fig. 8–49. (a) If each of the three rotor helicopter blades is 3.75 m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. (b) How much torque must the motor apply to bring the blades from rest to a speed of 6.0 rev/s in 8.0 s?

FIGURE 8-49

Problem 43

Calculate the moment of inertia of an align of point objects, as shown in Fig. 8–47 about (a) the y axis and (b) the x-axis. Assume two masses, \(m = 2.2\;{\rm{kg}}\)and \(M = 3.4\;{\rm{kg}}\), and the objects are wired together by very light, rigid pieces of wire. The align is rectangular and split through the middle by the x-axis. (c) About which axis would it be harder to accelerate this align?

FIGURE 8-47

Problem 39

Question:(II) A person stands, hands at his side, on a platform that is rotating at a rate of 0.90 rev/s. If he raises his arms to a horizontal position, Fig. 8–55, the speed of rotation decreases to 0.60 rev/s. (a) Why? (b) By what factor has his moment of inertia changed?

The moment of inertia of a rotating solid disk about an axis through its CM is \(\frac{{\bf{1}}}{{\bf{2}}}{\bf{M}}{{\bf{R}}^{\bf{2}}}\) (Fig. 8–20c). Suppose instead that a parallel axis of rotation passes through a point on the edge of the disk. Will the moment of inertia be the same, larger, or smaller? Explain why.

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