is attached to the midpoint of the string. The system is released from rest when the string is fully extended horizontally. Find the velocity of the mass at the instant the ring hits the hook. : Let the hook be the origin . Let the position of mass and the position of mass Set Constraints : The total length of the string is . The segment from the hook to has length . The segment from also has length Analyze the Boundary Condition : When the ring hits the hook, Determine the Geometry : Substituting into the second constraint gives
We want the block to stay still on the wedge.
, the equations of motion are derived systematically using the Euler-Lagrange equation: is attached to the midpoint of the string
200 Puzzling Physics Problems (P. Gnädig): Highly recommended for developing creative physical intuitions without relying on brute-force calculus.
ddt(𝜕L𝜕q̇i)−𝜕L𝜕qi=0d over d t end-fraction open paren the fraction with numerator partial cap L and denominator partial q dot sub i end-fraction close paren minus the fraction with numerator partial cap L and denominator partial q sub i end-fraction equals 0 Non-Inertial Reference Frames : Let the hook be the origin
T=12MẊ2+12m(ẋm2+ẏm2)+12Iω2cap T equals one-half cap M cap X dot squared plus one-half m open paren x dot sub m squared plus y dot sub m squared close paren plus one-half cap I omega squared For a solid cylinder, the moment of inertia is
T=12m(vθ2+vϕ2)=12mR2θ̇2+12mR2Ω2sin2θcap T equals one-half m open paren v sub theta squared plus v sub phi squared close paren equals one-half m cap R squared theta dot squared plus one-half m cap R squared cap omega squared sine squared theta The segment from also has length Analyze the
Using the equation: ΔU = mgh ΔU = 5(10)(10) = 500 J
ẋm2+ẏm2=(Ẋ+ẋcosα)2+(−ẋsinα)2=Ẋ2+2Ẋẋcosα+ẋ2x dot sub m squared plus y dot sub m squared equals open paren cap X dot plus x dot cosine alpha close paren squared plus open paren negative x dot sine alpha close paren squared equals cap X dot squared plus 2 cap X dot x dot cosine alpha plus x dot squared Thus, the total kinetic energy simplifies to:
Tomorrow, do another. In three months, you will see mechanics not as a series of formulas, but as an intuitive landscape of forces, energies, and symmetries. That is when the medals come.
U(x)=U0[(dx)2−2(dx)]cap U open paren x close paren equals cap U sub 0 open bracket open paren d over x end-fraction close paren squared minus 2 open paren d over x end-fraction close paren close bracket U0cap U sub 0 are positive constants. Find the equilibrium position of the particle. Determine if the equilibrium is stable. Calculate the angular frequency of small oscillations about this equilibrium point. Part 1: Finding Equilibrium