Question

A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel, of radius 0.357 m, and observes that drops of water fly off tangentially. She measures the height reached by drops moving vertically. A drop that breaks loose from the tire on one turn rises h = 54.2 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel.

Answer 1

1. Solve for the initial tangental velocity of both drops (V₀ and V₁ )

use: (V_final)² = (V_initia)_l² + 2 α Δr

V_final = will be zero when the drop reaches highest point. Solve for V_initial

V_{0 initial} = √(2 (9.8) (.542) = 3.26 m/s

V_{1 initial} = √(2 (9.8) (.510) = 3.16 m/s

2. Solve for ω (angular velocity):

ω = V /radius

ω₀ = 3.26 / .357 = 9.13 radians / second

ω₁ = 3.16 / .357 = 8.85 radians / second

3. Solve for α (angular acceleration) using circular motion kinematics:

(ω_final)² = (ω_initial)² + 2 α Δθ

8.85² = 9.13² + 2 α 2π

78.3-83.4 = 4π α

4πα = -5.1

α = -.406 radians/ second²

The magnitude of acceleration would be .406 radians/ second²