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Question-186439




Question Number 186439 by ajfour last updated on 04/Feb/23
Commented by ajfour last updated on 04/Feb/23
If inside the fixed hollow cone,  the solid ball of radius r revolves  sliding frictionlessly along the  lateral side and the base with  angular velocity ω, then find the  normal reaction forces it is  exerted upon by the ground and  the lateral side.
$${If}\:{inside}\:{the}\:{fixed}\:{hollow}\:{cone}, \\ $$$${the}\:{solid}\:{ball}\:{of}\:{radius}\:{r}\:{revolves} \\ $$$${sliding}\:{frictionlessly}\:{along}\:{the} \\ $$$${lateral}\:{side}\:{and}\:{the}\:{base}\:{with} \\ $$$${angular}\:{velocity}\:\omega,\:{then}\:{find}\:{the} \\ $$$${normal}\:{reaction}\:{forces}\:{it}\:{is} \\ $$$${exerted}\:{upon}\:{by}\:{the}\:{ground}\:{and} \\ $$$${the}\:{lateral}\:{side}. \\ $$
Answered by mr W last updated on 04/Feb/23
Commented by mr W last updated on 04/Feb/23
tan θ=(h/R)  b=R−(r/(tan (θ/2)))  F_C =mbω^2 =m(R−(r/(tan (θ/2))))ω^2   N_s  sin θ=F_C   N_s =(m/(sin θ))(R−(r/(tan (θ/2))))ω^2        =((m(√(h^2 +R^2 )))/h)(R−(r/( (√(1+(R^2 /h^2 )))−(R/h))))ω^2   N_b =mg+N_s cos θ       =mg+(m/(tan θ))(R−(r/(tan (θ/2))))ω^2        =mg+((mR)/h)(R−(r/( (√(1+(R^2 /h^2 )))−(R/h))))ω^2
$$\mathrm{tan}\:\theta=\frac{{h}}{{R}} \\ $$$${b}={R}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${F}_{{C}} ={mb}\omega^{\mathrm{2}} ={m}\left({R}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\omega^{\mathrm{2}} \\ $$$${N}_{{s}} \:\mathrm{sin}\:\theta={F}_{{C}} \\ $$$${N}_{{s}} =\frac{{m}}{\mathrm{sin}\:\theta}\left({R}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{{m}\sqrt{{h}^{\mathrm{2}} +{R}^{\mathrm{2}} }}{{h}}\left({R}−\frac{{r}}{\:\sqrt{\mathrm{1}+\frac{{R}^{\mathrm{2}} }{{h}^{\mathrm{2}} }}−\frac{{R}}{{h}}}\right)\omega^{\mathrm{2}} \\ $$$${N}_{{b}} ={mg}+{N}_{{s}} \mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:={mg}+\frac{{m}}{\mathrm{tan}\:\theta}\left({R}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:={mg}+\frac{{mR}}{{h}}\left({R}−\frac{{r}}{\:\sqrt{\mathrm{1}+\frac{{R}^{\mathrm{2}} }{{h}^{\mathrm{2}} }}−\frac{{R}}{{h}}}\right)\omega^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 04/Feb/23
pretty smooth n elegant solution.  Thank you sir!
$${pretty}\:{smooth}\:{n}\:{elegant}\:{solution}. \\ $$$${Thank}\:{you}\:{sir}! \\ $$

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