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Question Number 226136 by mr W last updated on 20/Nov/25
Commented by mr W last updated on 20/Nov/25
Commented by fantastic2 last updated on 20/Nov/25
i get (3/2)
$${i}\:{get}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 20/Nov/25
yes
$${yes} \\ $$
Answered by mr W last updated on 21/Nov/25
Commented by mr W last updated on 20/Nov/25
((mv^2 )/2)=mgR(1−cos θ) mg cos θ+N=((mv^2 )/R) N=mg(2−3 cos θ) 2N cos θ≥Mg 2mg cos θ (2−3 cos θ)≥Mg (m/M)≥(1/(2 cos θ (2−3 cos θ)))≥(3/2) minimum is at cos θ=(2/3)−cos θ, i.e. cos θ=(1/3)
$$\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${mg}\:\mathrm{cos}\:\theta+{N}=\frac{{mv}^{\mathrm{2}} }{{R}} \\ $$$${N}={mg}\left(\mathrm{2}−\mathrm{3}\:\mathrm{cos}\:\theta\right) \\ $$$$\mathrm{2}{N}\:\mathrm{cos}\:\theta\geqslant{Mg} \\ $$$$\mathrm{2}{mg}\:\mathrm{cos}\:\theta\:\left(\mathrm{2}−\mathrm{3}\:\mathrm{cos}\:\theta\right)\geqslant{Mg} \\ $$$$\frac{{m}}{{M}}\geqslant\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\theta\:\left(\mathrm{2}−\mathrm{3}\:\mathrm{cos}\:\theta\right)}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${minimum}\:{is}\:{at}\: \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{cos}\:\theta,\:{i}.{e}.\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by fantastic2 last updated on 21/Nov/25
N+mgcos θ=mω^2 R=m(v^2 /R) v^2 =((NR+mgRcos θ)/m) ....i (1/2)mv^2 +mgRcos θ=mgR v^2 =2gR(1−cos θ)....ii i=ii 2mg(1−cos θ)=N+mgcos θ N=mg(2−3cos θ) upward vertical F N_y =Ncos θ F_(total) =2Ncos θ=2mg(2−3cos θ)cos θ at F_(total_(max) ) (dF/dθ)=0 2mg(d/dθ)(2−3cos θ)cos θ=0 ⇒6sin θcos θ−2sin θ=0 ⇒3cos θ=1 cos θ=(1/3) F_y_(max) =(2/3)mg F_y_(max) ≥Mg (m/M)≥(3/2)
$${N}+{mg}\mathrm{cos}\:\theta={m}\omega^{\mathrm{2}} {R}={m}\frac{{v}^{\mathrm{2}} }{{R}} \\ $$$${v}^{\mathrm{2}} =\frac{{NR}+{mgR}\mathrm{cos}\:\theta}{{m}}\:….{i} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +{mgR}\mathrm{cos}\:\theta={mgR} \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)….{ii} \\ $$$${i}={ii} \\ $$$$\mathrm{2}{mg}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)={N}+{mg}\mathrm{cos}\:\theta \\ $$$${N}={mg}\left(\mathrm{2}−\mathrm{3cos}\:\theta\right) \\ $$$${upward}\:{vertical}\:{F}\: \\ $$$${N}_{{y}} ={N}\mathrm{cos}\:\theta \\ $$$${F}_{{total}} =\mathrm{2}{N}\mathrm{cos}\:\theta=\mathrm{2}{mg}\left(\mathrm{2}−\mathrm{3cos}\:\theta\right)\mathrm{cos}\:\theta \\ $$$${at}\:{F}_{{total}_{{max}} } \frac{{dF}}{{d}\theta}=\mathrm{0} \\ $$$$\mathrm{2}{mg}\frac{{d}}{{d}\theta}\left(\mathrm{2}−\mathrm{3cos}\:\theta\right)\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{6sin}\:\theta\mathrm{cos}\:\theta−\mathrm{2sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${F}_{{y}_{{max}} } =\frac{\mathrm{2}}{\mathrm{3}}{mg} \\ $$$${F}_{{y}_{{max}} } \geqslant{Mg} \\ $$$$\frac{{m}}{{M}}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 21/Nov/25
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Commented by fantastic2 last updated on 21/Nov/25
Q226143. am i?right sir
$${Q}\mathrm{226143}.\:{am}\:{i}?{right}\:{sir} \\ $$

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