vf=v0−μg(2v07μg)=v0−27v0=57v0v sub f equals v sub 0 minus mu g of open paren the fraction with numerator 2 v sub 0 and denominator 7 mu g end-fraction close paren equals v sub 0 minus two-sevenths v sub 0 equals five-sevenths v sub 0
above the vertex at which the block can remain stationary relative to the rotating cone. If you'd like, I can:
Search for to find advanced classical mechanics deep dives. If you'd like
d2Ueffdθ2|θ=θ0=Mg2ω2−2Mg2ω2+Mω2R2=Mω2R2(1−g2ω4R2)the fraction with numerator d squared cap U sub e f f end-sub and denominator d theta squared end-fraction vertical line sub theta equals theta sub 0 end-sub equals the fraction with numerator cap M g squared and denominator omega squared end-fraction minus the fraction with numerator 2 cap M g squared and denominator omega squared end-fraction plus cap M omega squared cap R squared equals cap M omega squared cap R squared open paren 1 minus the fraction with numerator g squared and denominator omega to the fourth power cap R squared end-fraction close paren Since this configuration requires If you'd like, I can:
a=−fkM=−μga equals the fraction with numerator negative f sub k and denominator cap M end-fraction equals negative mu g
When reviewing the linked solutions, focus on why a particular method was chosen rather than just the math. If you'd like, I can: