Question-222961

Question Number 222961 by Jubr last updated on 12/Jul/25

Answered by mr W last updated on 12/Jul/25

Commented by mr W last updated on 12/Jul/25

f=μN=μm(g cos θ+(v^2 /R)) =0.5×1×(9.81×cos 60°+(4^2 /2))≈6.45 N

$${f}=\mu{N}=\mu{m}\left({g}\:\mathrm{cos}\:\theta+\frac{{v}^{\mathrm{2}} }{{R}}\right) \\ $$$$\:\:=\mathrm{0}.\mathrm{5}×\mathrm{1}×\left(\mathrm{9}.\mathrm{81}×\mathrm{cos}\:\mathrm{60}°+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}\right)\approx\mathrm{6}.\mathrm{45}\:{N} \\ $$

Commented by Jubr last updated on 12/Jul/25

$${Thanks}\:{sir}. \\ $$

Commented by Tawa11 last updated on 12/Jul/25

Weldone sir. Please make the question very complex. I want to try the complex part.

$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{make}\:\mathrm{the}\:\mathrm{question}\:\mathrm{very}\:\mathrm{complex}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{try}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{part}. \\ $$

Commented by mr W last updated on 12/Jul/25

Commented by mr W last updated on 13/Jul/25

a small object of mass m is released from rest at θ=60° on the smooth circular track with radius R. find the period T of the motion of the object.

$${a}\:{small}\:{object}\:{of}\:{mass}\:\boldsymbol{{m}}\:{is}\:{released} \\ $$$${from}\:{rest}\:{at}\:\theta=\mathrm{60}°\:{on}\:{the}\:{smooth} \\ $$$${circular}\:{track}\:{with}\:{radius}\:\boldsymbol{{R}}.\: \\ $$$${find}\:{the}\:{period}\:\boldsymbol{{T}}\:{of}\:{the}\:{motion}\: \\ $$$${of}\:{the}\:{object}. \\ $$

Commented by mr W last updated on 17/Jul/25

v=Rω ((mv^2 )/2)=mgR(cos θ−cos θ_0 ) ω^2 =((2g(cos θ−cos θ_0 ))/R) −(dθ/dt)=ω=(√((2g(cos θ−cos θ_0 ))/R)) −(√(R/(2g)))∫_θ_0 ^0 (dθ/( (√(cos θ−cos θ_0 ))))=∫_0 ^(T/4) dt T=4(√(R/(2g)))∫_0 ^θ_0 (dθ/( (√(cos θ−cos θ_0 ))))=4λ(√(R/(2g))) with λ=∫_0 ^θ_0 (dθ/( (√(cos θ−cos θ_0 )))) for θ_0 =60°: λ≈2.38401101 T≈6.743(√(R/g))

$${v}={R}\omega \\ $$$$\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}={mgR}\left(\mathrm{cos}\:\theta−\mathrm{cos}\:\theta_{\mathrm{0}} \right) \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}\left(\mathrm{cos}\:\theta−\mathrm{cos}\:\theta_{\mathrm{0}} \right)}{{R}} \\ $$$$−\frac{{d}\theta}{{dt}}=\omega=\sqrt{\frac{\mathrm{2}{g}\left(\mathrm{cos}\:\theta−\mathrm{cos}\:\theta_{\mathrm{0}} \right)}{{R}}} \\ $$$$−\sqrt{\frac{{R}}{\mathrm{2}{g}}}\int_{\theta_{\mathrm{0}} } ^{\mathrm{0}} \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta−\mathrm{cos}\:\theta_{\mathrm{0}} }}=\int_{\mathrm{0}} ^{\frac{{T}}{\mathrm{4}}} {dt} \\ $$$${T}=\mathrm{4}\sqrt{\frac{{R}}{\mathrm{2}{g}}}\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta−\mathrm{cos}\:\theta_{\mathrm{0}} }}=\mathrm{4}\lambda\sqrt{\frac{{R}}{\mathrm{2}{g}}} \\ $$$${with}\:\lambda=\int_{\mathrm{0}} ^{\theta_{\mathrm{0}} } \frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta−\mathrm{cos}\:\theta_{\mathrm{0}} }} \\ $$$${for}\:\theta_{\mathrm{0}} =\mathrm{60}°:\: \\ $$$$\lambda\approx\mathrm{2}.\mathrm{38401101} \\ $$$${T}\approx\mathrm{6}.\mathrm{743}\sqrt{\frac{{R}}{{g}}} \\ $$

Commented by Tawa11 last updated on 18/Jul/25

$$\mathrm{Ohh}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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