Question

A bicycle racer is going downhill at 12.0 m/s when, to his horror, one of his 12.5kg wheel comes off when he is at top of the hill. We can model the wheel as a thick-walled cylinder of outer diameter 0.85 m, with a moment of inertia of 0.367 kg-m². (Neglect the small mass of the spokes). Consider its motion is rolling without slipping, and downhill angular acceleration of 2.3 rad/s². What distance has the wheel traveled, in meters, when the wheel is going at 29.3 m/s?

Answer 1

To find the distance traveled by a wheel rolling downhill from an initial speed of 12.0 m/s to 29.3 m/s with an angular acceleration of 2.3 rad/s², we use kinematics and rotational dynamics to calculate a distance of approximately 430.17 meters.

Step-by-step explanation:

The bicycle racer's wheel is described as a thick-walled cylinder with known moment of inertia and rolling downhill without slipping. To solve for the distance traveled when the wheel reaches a certain speed, we can use kinematic equations and rotational dynamics. The wheel starts at a translational velocity of 12.0 m/s and reaches 29.3 m/s, with an angular acceleration of 2.3 rad/s².

First, we determine the angular velocity at the final speed using the relationship between translational velocity (v) and angular velocity (ω): v = r × ω. With an outer diameter of 0.85 m, the radius r is 0.425 m. Thus, ω = v/r = 29.3 / 0.425 = 68.941 rad/s. Then, we calculate the initial angular velocity with the same formula using the initial translational velocity: ω = 12.0 / 0.425 = 28.235 rad/s.

Now, using the kinematic equation ω_f = ω_i + α × t, we solve for time t: t = (ω_f - ω_i) / α = (68.941 - 28.235) / 2.3 = 17.695 s. Using the final time and the linear acceleration a = r × α = 0.425 × 2.3 = 0.9775 m/s², we calculate the distance using s = v_i × t + ½ × a × t². The total distance covered is s = 12.0 × 17.695 + ½ × 0.9775 × (17.695)² = 430.17 m (approx.).