Uningenia
Start studyingStart
KinematicsHard

Circular motion with variable acceleration

A point moves along a circle of radius R = 0.20 m, initially with angular velocity ω₀ = 0.5 rad/s. At t = 0 it accelerates with α = 0.15·t rad/s³. At t = 6 s find: angular velocity, total acceleration and the angle with the tangent.
R aₙ aₜ ω
Given data
R = 0.20 mω₀ = 0.5 rad/sα(t) = 0.15·t rad/s³t = 6 s
Review the theory: Cinematica
Steps
0 / 5
  1. Angular velocity grows because there is an angular acceleration α. But α depends on time, so you can't use ω = ω₀ + αt (valid only if α is constant): you must integrate α over time. ω(t) = ω₀ + ∫₀ᵗ 0.15t dt = ω₀ + 0.075·t².
  2. Even if the speed were constant in magnitude, the point curves: the direction of the velocity keeps changing. This requires an acceleration pointing toward the center, the centripetal one: aₙ = ω²·R.
  3. Here the speed is also increasing in magnitude: this change is the tangential acceleration, directed along the path. It is linked to the angular acceleration by aₜ = R·α(t).
  4. aₙ (radial) and aₜ (tangential) are perpendicular, so the total acceleration is their vector sum: the magnitude comes from the Pythagorean theorem, a = √(aₜ² + aₙ²).
  5. The angle between total acceleration and the tangent measures how much the acceleration points toward the center. From the ratio of radial to tangential components: θ = arctan(aₙ/aₜ).
Full worked solution
  1. Angular velocity grows because there is an angular acceleration α. But α depends on time, so you can't use ω = ω₀ + αt (valid only if α is constant): you must integrate α over time. ω(t) = ω₀ + ∫₀ᵗ 0.15t dt = ω₀ + 0.075·t².
    ω=ω0+0.075 t2=0.5+0.075⋅62\omega = \omega_0 + 0.075\,t^2 = 0.5 + 0.075\cdot 6^2
    ω=0.5+0.075⋅36=3.2 rad/s\omega = 0.5 + 0.075\cdot36 = \mathbf{3.2\,rad/s}.
  2. Even if the speed were constant in magnitude, the point curves: the direction of the velocity keeps changing. This requires an acceleration pointing toward the center, the centripetal one: aₙ = ω²·R.
    an=ω2R=3.22⋅0.20a_n = \omega^2 R = 3.2^2\cdot 0.20
    an=10.24⋅0.20=2.05 m/s2a_n = 10.24\cdot0.20 = \mathbf{2.05\,m/s^2}.
  3. Here the speed is also increasing in magnitude: this change is the tangential acceleration, directed along the path. It is linked to the angular acceleration by aₜ = R·α(t).
    at=R α=0.20⋅0.15⋅6a_t = R\,\alpha = 0.20\cdot 0.15\cdot 6
    at=0.20⋅0.9=0.18 m/s2a_t = 0.20\cdot0.9 = \mathbf{0.18\,m/s^2}.
  4. aₙ (radial) and aₜ (tangential) are perpendicular, so the total acceleration is their vector sum: the magnitude comes from the Pythagorean theorem, a = √(aₜ² + aₙ²).
    a=at2+an2a = \sqrt{a_t^2 + a_n^2}
    a=0.032+4.20≈2.06 m/s2a = \sqrt{0.032+4.20}\approx\mathbf{2.06\,m/s^2} (almost entirely centripetal).
  5. The angle between total acceleration and the tangent measures how much the acceleration points toward the center. From the ratio of radial to tangential components: θ = arctan(aₙ/aₜ).
    θ=arctan⁡ ⁣(anat)\theta = \arctan\!\left(\frac{a_n}{a_t}\right)
    θ=arctan⁡(11.4)≈85∘\theta = \arctan(11.4)\approx\mathbf{85^\circ}: acceleration almost radial.
Result:ω=3.2\omega = 3.2 rad/s, a≈2.06a \approx 2.06 m/s², θ≈85°\theta \approx 85°
TheoryCinematicaFormula sheetPhysics 1

Similar exercises