: Initial energy is purely potential. If we set the initial horizontal line as Ei=0cap E sub i equals 0 At the final instant, mass has dropped by a vertical distance , and mass has dropped by (it returns to the origin hook).
=−(u−vcosθ)+vu(ucosθ)=−u+vcosθ+vcosθ (This doesn′t cancel directly)equals negative open paren u minus v cosine theta close paren plus v over u end-fraction open paren u cosine theta close paren equals negative u plus v cosine theta plus v cosine theta (This doesn prime t cancel directly) Instead, let's look at the quantity
Using Newton's second law: F - f = ma 10 - f = 2(3) f = 4 N
. Find the equilibrium positions of the bead and analyze their stability as a function of
https://www.ipho.org/problems-solutions This is the holy grail. The official IPhO archive contains every problem and solution from 1967 to the present. Problems are presented in English and the official working language. Why use it? Authenticity. If you can solve the last 10 IPhO mechanics problems (e.g., “Spinning Cylinder on a Table” or “Collision of Galaxies”), you are ready for any national team selection camp. : Initial energy is purely potential
We find the equilibrium states by looking at the effective potential energy
. Find the acceleration of the plank and the cylinder if slipping occurs. : The plank experiences the external force forward and a friction force
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: The official archive for the International Physics Olympiad , featuring theoretical and experimental mechanics problems from 1967 to the present. Find the equilibrium positions of the bead and
is placed on the incline of the wedge. Both are released from rest. Find the of the wedge ( ) relative to the floor. The Solution
. The potential energy is at a local minimum, meaning the equilibrium is . Part 3: Frequency of Small Oscillations For small displacements
: Slipping occurs when the required friction exceeds the maximum static friction ( ). Under kinetic slipping, the friction force is exactly: f=μmgf equals mu m g Calculate Accelerations : Substitute into the cylinder's linear equation:
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This is a linear differential equation with a modulating term. Let's make a substitution to eliminate the first derivative term (transforming to a rotating frame). Let .Differentiating
12xddx(x2v2)=Fμ1 over 2 x end-fraction d over d x end-fraction open paren x squared v squared close paren equals the fraction with numerator cap F and denominator mu end-fraction