: 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).
ddt[r+vu(xA−x)]=(−u+vcosθ)+(v2u−vcosθ)=−u+v2u=−u2−v2ud over d t end-fraction open bracket r plus v over u end-fraction open paren x sub cap A minus x close paren close bracket equals open paren negative u plus v cosine theta close paren plus open paren the fraction with numerator v squared and denominator u end-fraction minus v cosine theta close paren equals negative u plus the fraction with numerator v squared and denominator u end-fraction equals negative the fraction with numerator u squared minus v squared and denominator u end-fraction This is a constant! We can integrate this directly from (the time of interception). Therefore,
Using the equation: f = μN 4 = μ(2)(10) μ = 0.2 : Initial energy is purely potential
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Notice that the left side of the equation is a perfect derivative: We can integrate this directly from (the time
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that rotates about its vertical diameter with a constant angular velocity Notice that the left side of the equation
:
. The potential energy is at a local minimum, meaning the equilibrium is . Part 3: Frequency of Small Oscillations For small displacements
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Assuming a mass of 1 kg for the wheel: