ASVAB (Armed Services Vocational Aptitude Battery)Mechanical Comprehension (MC)Hard
A bicycle wheel has a diameter of 0.6 meters. If the wheel completes 10 revolutions per second, what is the approximate linear speed of a point on the outer edge of the wheel?
- A3.14 m/s
- B18.85 m/s
- C1.88 m/s
- D6.28 m/s
Show answer & explanationAnswer & explanation
Correct answer: B. 18.85 m/s
The linear speed (v) of a point on the edge of a rotating object is given by v = ωr, where ω is the angular velocity in radians per second and r is the radius. First, calculate the radius: r = diameter / 2 = 0.6 m / 2 = 0.3 m. Next, calculate angular velocity: 1 revolution = 2π radians. So, ω = 10 revolutions/second × 2π radians/revolution = 20π rad/s. Finally, v = (20π rad/s) × 0.3 m = 6π m/s ≈ 6 × 3.14159 m/s ≈ 18.85 m/s.
Why the other options are wrong
- A. Incorrect, this is π m/s, or 1 revolution per second with radius 0.5m.
- C. Incorrect, likely a calculation error or used radius as diameter.
- D. Incorrect, this is 2π m/s, or 1 revolution per second with radius 1m.
Linear Speed from Angular Speed
The linear speed of a point on a rotating object is directly proportional to its distance from the center of rotation and its angular speed.
- Formula: v = ωr.
- ω (angular speed) must be in radians per second (rad/s).
- r (radius) must be in meters (m).
- Linear speed (v) is in meters per second (m/s).
Memory trick: Angular 'omega' times 'radius' gives 'v' linear.