Bead on a Rotating Hoop: Wrong Sign, Wrong Equilibrium
A bead of mass m slides without friction on a circular hoop of radius R = 0.50 m. The hoop rotates about its vertical diameter at constant angular speed ω. In a lab demonstration ω = 6.0 rad/s and g = 9.8 m/s². A student writes the effective potential as V_eff(θ) = mgR(1 - cosθ) + (1/2)mω²R² sin²θ, linearizes about the bottom θ=0, and predicts a stable equilibrium with frequency sqrt(g/R + ω²R) ≈ 6.1 rad/s. In the demo the bead does not stay at the bottom; it settles at a nonzero angle. You are asked to correct the sign of the centrifugal term, determine the actual equilibrium angle, and compute the small-oscillation frequency about it. Treat the bead as a point mass and ignore the hoop's own mass.
1. Write the correct effective potential and find all equilibrium angles for ω = 6.0 rad/s.
Read the worked answer
The centrifugal term lowers the effective potential away from the axis, so the correct expression is
V_eff(θ) = mgR(1 - cosθ) - (1/2)mω²R² sin²θ.
Differentiating,
V_eff'(θ) = mR sinθ (g - ω²R cosθ).
Equilibria require sinθ = 0 or cosθ = g/(ω²R). Here ω²R = 36 × 0.50 = 18 m/s², so g/(ω²R) = 9.8/18 = 0.5444. Therefore θ = 0, θ = π, and θ = ±arccos(0.5444) ≈ ±57.0°. The angle θ = π is not an additional nontrivial equilibrium because cosθ = -1 does not satisfy the second condition; it is just the top point from sinθ = 0.
2. Classify the stability of θ = 0 and of the nonzero equilibrium.
Read the worked answer
The second derivative is
V_eff''(θ) = mR[g cosθ - ω²R cos2θ].
At θ = 0, cosθ = 1 and cos2θ = 1, so
V_eff''(0) = mR(g - ω²R) = m(0.50)(9.8 - 18) < 0.
The bottom is therefore unstable, not stable. At θ0 with cosθ0 = g/(ω²R), use cos2θ0 = 2cos²θ0 - 1. Substitution gives
V_eff''(θ0) = mR[ω²R - g²/(ω²R)] > 0.
Numerically, ω²R = 18 and g²/(ω²R) = 96.04/18 ≈ 5.34, so the bracket is positive. The nonzero angle is a stable equilibrium. The student's sign error turned the centrifugal term into a restoring term and removed the true instability at the bottom.
3. Compute the small-oscillation frequency about θ0 and compare with the student's 6.1 rad/s.
Read the worked answer
The kinetic energy is T = (1/2)mR² θdot², so the angular frequency about θ0 is
Ω² = V_eff''(θ0)/(mR²) = ω² - g²/(ω²R²).
With the given numbers,
Ω² = 36 - 96.04/(36 × 0.25) = 36 - 10.67 = 25.33 s⁻²,
so Ω ≈ 5.03 rad/s. The student's 6.1 rad/s comes from the wrong-sign potential and has no physical meaning for this operating point. The correct motion is a small oscillation about the cone angle θ0 ≈ 57°, not about the bottom. If ω were reduced below sqrt(g/R) ≈ 4.43 rad/s, the nonzero equilibrium would disappear and θ = 0 would become stable; the demonstration is on the other side of that bifurcation.