slides without friction on a vertical circular wire loop of radius
Mg−T=Ma(Equation 1)cap M g minus cap T equals cap M a space (Equation 1) Next, we apply the rotational form of Newton's Second Law (
For focused preparation, use these collections that feature both challenging problems and detailed solutions: Jaan Kalda’s Mechanics Guide
that rotates about its vertical diameter with a constant angular velocity slides without friction on a vertical circular wire
If a link changes, Google: "F=ma 2022 solutions PDF" or "USAPhO 2019 problem 2 solution" .
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
. Assuming the amplitude of oscillation is small and neglecting higher-order terms of the Earth's angular velocity Ωcap omega , derive the angular velocity ωpomega sub p Assuming the amplitude of oscillation is small and
Use energy conservation and set normal force (N = 0) → (\cos\theta = 2/3). Correct solution insight: You must also account that the block’s radial acceleration is (v^2/R) and (N = mg\cos\theta - m v^2/R). The instant (N=0) gives (v^2 = gR\cos\theta). Combine with energy (mgR(1-\cos\theta) = \frac12 mv^2) → (\cos\theta = 2/3).
(Simulated Link)
| Step | Action | |------|--------| | 1 | Start with or F=ma past exams – build speed and accuracy. | | 2 | Move to Irodov selected problems (e.g., dynamics of rigid bodies). | | 3 | Study Morin’s book excerpts for unconventional mechanical reasoning. | | 4 | Attempt IPhO official mechanics problems (years 2015–present). | | 5 | Simulate contest: solve USAPhO semifinal problems under time limit, then check against official solutions. | Combine with energy (mgR(1-\cos\theta) = \frac12 mv^2) →
where r is the radius and F is the force.
𝜕2Veff𝜕θ2|θ=0=mgR−mR2Ω2=mR2(gR−Ω2)the fraction with numerator partial squared cap V sub e f f end-sub and denominator partial theta squared end-fraction vertical line sub theta equals 0 end-sub equals m g cap R minus m cap R squared cap omega squared equals m cap R squared open paren the fraction with numerator g and denominator cap R end-fraction minus cap omega squared close paren For stability, this value must be greater than zero:
[r+vu(xA−x)]t=0=dopen bracket r plus v over u end-fraction open paren x sub cap A minus x close paren close bracket sub t equals 0 end-sub equals d intercepts Therefore,