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Disk rolling on an inclined plane

A homogeneous disk (M = 0.5 kg, R = 5 cm) rolls without slipping on a plane inclined at θ = 25° (μs = 0.4). Find: angular acceleration, friction force, and the maximum angle before slipping.
mg N f θ
Given data
M = 0.5 kgR = 0.05 mθ = 25°μs = 0.4
Review the theory: Dinamica
Steps
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  1. The disk rolls without slipping, so the constraint a = R·α holds. Two equations are needed: (1) translation along the plane → Mg·sinθ − F_a = M·a; (2) rotation about the center → F_a·R = I·α, with I = ½MR² (disk). From (2), F_a = ½M·R·α; substituting it into (1) together with a = R·α, friction cancels and α = 2g·sinθ/(3R) remains.
  2. It is precisely static friction that supplies the torque making the disk roll (without friction it would slide without rotating). I first find the linear acceleration a = R·α, then from the 2nd law along the plane Mg·sinθ − F_a = M·a I get the friction force F_a.
  3. Static friction has a maximum value μs·N. The steeper the plane, the more friction pure rolling needs: beyond a certain angle the maximum friction is not enough and the disk starts to slip. Imposing required F_a ≤ μs·N gives the limiting condition θ_max = arctan(3μs).
Full worked solution
  1. The disk rolls without slipping, so the constraint a = R·α holds. Two equations are needed: (1) translation along the plane → Mg·sinθ − F_a = M·a; (2) rotation about the center → F_a·R = I·α, with I = ½MR² (disk). From (2), F_a = ½M·R·α; substituting it into (1) together with a = R·α, friction cancels and α = 2g·sinθ/(3R) remains.
    α=2gsin⁡θ3R=2⋅9.81⋅sin⁡25°3⋅0.05\alpha = \frac{2g\sin\theta}{3R} = \frac{2\cdot9.81\cdot\sin25°}{3\cdot0.05}
    α=8.290.15≈55.3 rad/s2\alpha = \frac{8.29}{0.15} \approx \mathbf{55.3\,rad/s^2}.
  2. It is precisely static friction that supplies the torque making the disk roll (without friction it would slide without rotating). I first find the linear acceleration a = R·α, then from the 2nd law along the plane Mg·sinθ − F_a = M·a I get the friction force F_a.
    Fa=Mgsin⁡θ−MRαF_a = Mg\sin\theta - M R\alpha
    a=2.76 m/s2a = 2.76\,m/s^2; Fa=2.07−1.38=0.69 NF_a = 2.07 - 1.38 = \mathbf{0.69\,N}.
  3. Static friction has a maximum value μs·N. The steeper the plane, the more friction pure rolling needs: beyond a certain angle the maximum friction is not enough and the disk starts to slip. Imposing required F_a ≤ μs·N gives the limiting condition θ_max = arctan(3μs).
    θmax=arctan⁡(3μs)=arctan⁡(1.2)\theta_{max} = \arctan(3\mu_s) = \arctan(1.2)
    θmax=arctan⁡(1.2)≈50.2∘\theta_{max} = \arctan(1.2) \approx \mathbf{50.2^\circ}.
Result:α≈55.3\alpha \approx 55.3 rad/s², Fa≈0.69F_a \approx 0.69 N, θmax≈50.2°\theta_{max} \approx 50.2°
TheoryDinamicaFormula sheetPhysics 1

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