A wheel 2.0 m in diameter lies in the vertical plane and rotates about its central axis with a constant angular acceleration of 4.0 rad s⁻². The wheel starts at rest at t = 0 and the radius vector of a point A on the wheel makes an angle of 60º with the horizontal at this instant. Calculate the angular speed of the wheel, the angular position of the point A and the total acceleration at t = 2.0s.

Introduction

This problem involves rotational motion with constant angular acceleration. We’re asked to find the angular speed, angular position, and total acceleration of a point on the rim of a rotating wheel after 2 seconds.

Given Data

Diameter of wheel = 2.0 m ⇒ Radius (r) = 1.0 m
Angular acceleration (α) = 4.0 rad/s²
Initial angular velocity (ω₀) = 0 (starts from rest)
Time, t = 2.0 s
Initial angle between radius vector of point A and horizontal = 60º

Step 1: Angular Speed at t = 2.0 s

Using the formula for angular velocity under constant angular acceleration:

ω = ω₀ + αt

ω = 0 + (4.0)(2.0) = 8.0 rad/s

Step 2: Angular Position (θ) at t = 2.0 s

Using the formula for angular displacement:

θ = ω₀t + ½αt²

θ = 0 + 0.5 × 4 × (2)² = 0.5 × 4 × 4 = 8.0 radians

This is the angle turned by the wheel from t = 0.

Since 1 revolution = 2π ≈ 6.28 rad, the wheel turns a little over 1 revolution.

Total angular position of point A = Initial angle + rotation

Initial angle = 60º = π/3 rad ≈ 1.047 rad

Final angular position = 1.047 + 8.0 = 9.047 rad

We can convert this to an equivalent angle within one circle by subtracting multiples of 2π:

9.047 – 6.283 = 2.764 rad (remaining angle after 1 full rotation)

In degrees: (2.764 × 180) / π ≈ 158.4º from horizontal

Step 3: Total Acceleration

Total acceleration of a rotating point has two components:

Tangential acceleration (a_t) = r × α = 1.0 × 4.0 = 4.0 m/s²
Centripetal (radial) acceleration (a_c) = r × ω² = 1.0 × (8.0)² = 64 m/s²

Total acceleration (a_total) is the vector sum:

a_total = √(a_t² + a_c²) = √(4² + 64²) = √(16 + 4096) = √4112 ≈ 64.1 m/s²

Final Answers

Angular speed at t = 2s: 8.0 rad/s
Angular position of point A: ~158.4º from horizontal
Total acceleration: ~64.1 m/s²

Conclusion

This problem involves application of basic rotational kinematics. By calculating angular speed, angular displacement, and combining tangential and centripetal accelerations, we find that the wheel reaches an angular speed of 8.0 rad/s and the point A on its edge has a total acceleration of around 64.1 m/s² after 2 seconds.