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Question-225506

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Question Number 225506 by fantastic last updated on 31/Oct/25
Commented by fantastic last updated on 31/Oct/25
youre right sir
$${youre}\:{right}\:{sir} \\ $$
Commented by MirHasibulHossain last updated on 01/Nov/25
where are you from?
$$\mathrm{where}\:\mathrm{are}\:\mathrm{you}\:\mathrm{from}? \\ $$
Answered by mr W last updated on 31/Oct/25
Commented by mr W last updated on 31/Oct/25
N_2 =f_1 =kN_1 mg=N_1 +f_2 =N_1 +kN_2 =N_1 +k^2 N_1 ⇒N_1 =((mg)/(1+k^2 )) ⇒f_1 =((kmg)/(1+k^2 )), f_2 =((k^2 mg)/(1+k^2 )) τ=(f_1 +f_2 )R=((k(1+k)mgR)/(1+k^2 )) ((mR^2 )/2)α=τ=((k(1+k)mgR)/(1+k^2 )) ⇒α=((2k(1+k)g)/((1+k^2 )R)) θ=(ω_0 ^2 /(2α))=(((1+k^2 )Rω_0 ^2 )/(4k(1+k)g)) ⇒n=(θ/(2π))=(((1+k^2 )Rω_0 ^2 )/(8πk(1+k)g)) ✓
$${N}_{\mathrm{2}} ={f}_{\mathrm{1}} ={kN}_{\mathrm{1}} \\ $$$${mg}={N}_{\mathrm{1}} +{f}_{\mathrm{2}} ={N}_{\mathrm{1}} +{kN}_{\mathrm{2}} ={N}_{\mathrm{1}} +{k}^{\mathrm{2}} {N}_{\mathrm{1}} \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{{mg}}{\mathrm{1}+{k}^{\mathrm{2}} }\: \\ $$$$\Rightarrow{f}_{\mathrm{1}} =\frac{{kmg}}{\mathrm{1}+{k}^{\mathrm{2}} },\:{f}_{\mathrm{2}} =\frac{{k}^{\mathrm{2}} {mg}}{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\tau=\left({f}_{\mathrm{1}} +{f}_{\mathrm{2}} \right){R}=\frac{{k}\left(\mathrm{1}+{k}\right){mgR}}{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\alpha=\tau=\frac{{k}\left(\mathrm{1}+{k}\right){mgR}}{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}{k}\left(\mathrm{1}+{k}\right){g}}{\left(\mathrm{1}+{k}^{\mathrm{2}} \right){R}} \\ $$$$\theta=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\alpha}=\frac{\left(\mathrm{1}+{k}^{\mathrm{2}} \right){R}\omega_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}{k}\left(\mathrm{1}+{k}\right){g}} \\ $$$$\Rightarrow{n}=\frac{\theta}{\mathrm{2}\pi}=\frac{\left(\mathrm{1}+{k}^{\mathrm{2}} \right){R}\omega_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{8}\pi{k}\left(\mathrm{1}+{k}\right){g}}\:\checkmark \\ $$
Commented by ajfour last updated on 02/Nov/25
https://youtu.be/p-g2WKPFhXE?si=BjwljtbJIyClXtjt
Commented by fantastic last updated on 02/Nov/25
https://youtu.be/cAG1ijYC_y0?si=sbHvuRy60vhkW2oS
Commented by fantastic last updated on 02/Nov/25
(1+R)^2 =R^2 +(R−1)^2 R^2 =4R R=4
$$\left(\mathrm{1}+{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left({R}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\mathrm{4}{R} \\ $$$${R}=\mathrm{4} \\ $$
Commented by ajfour last updated on 02/Nov/25
��what i shared is a better question. Your shared one is damn easy!
Commented by fantastic last updated on 02/Nov/25
or (1,0) and (R,R) (√((1−R)^2 +(0−R)^2 ))=1+R R^2 =4R R⇒4
$${or} \\ $$$$\left(\mathrm{1},\mathrm{0}\right)\:{and}\:\left({R},{R}\right) \\ $$$$\sqrt{\left(\mathrm{1}−{R}\right)^{\mathrm{2}} +\left(\mathrm{0}−{R}\right)^{\mathrm{2}} }=\mathrm{1}+{R} \\ $$$${R}^{\mathrm{2}} =\mathrm{4}{R} \\ $$$${R}\Rightarrow\mathrm{4} \\ $$
Commented by fantastic last updated on 02/Nov/25
yes
$${yes} \\ $$
Commented by fantastic last updated on 02/Nov/25
i havent seen your full video. i will try to give my best will also share my approach/solution
$${i}\:{havent}\:{seen}\:{your}\:{full}\:{video}. \\ $$$${i}\:{will}\:{try}\:{to}\:{give}\:{my}\:{best} \\ $$$${will}\:{also}\:{share}\:{my}\:{approach}/{solution} \\ $$
Commented by fantastic last updated on 02/Nov/25
Commented by fantastic last updated on 02/Nov/25
R^2 +x^2 =(R+1)^2 x=(√(1+2R)) (1+(√(1+2R)))^2 +(R−(√3))^2 =(R+1)^2 1+1+2R+2(√(1+2R))+R^2 +3−2(√3)R=R^2 +1+2R 4+2(√(1+2R))=2(√3)R 2+(√(1+2R))=(√3)R 1+2R=3R^2 +4−4(√3)R 3R^2 −R(4(√3)+2)+3=0 R=(((4(√3)+2)±(√(48+4+16(√3)−36)))/6) =((2(√3)+1)/3)±((2(√((1+(√3)))))/3)
$${R}^{\mathrm{2}} +{x}^{\mathrm{2}} =\left({R}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}=\sqrt{\mathrm{1}+\mathrm{2}{R}} \\ $$$$\left(\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2}{R}}\right)^{\mathrm{2}} +\left({R}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\left({R}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\cancel{\mathrm{1}}+\cancel{\mathrm{2}{R}}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{2}{R}}+\cancel{{R}^{\mathrm{2}} }+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}{R}=\cancel{{R}^{\mathrm{2}} }+\cancel{\mathrm{1}}+\cancel{\mathrm{2}{R}} \\ $$$$\mathrm{4}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{2}{R}}=\mathrm{2}\sqrt{\mathrm{3}}{R} \\ $$$$\mathrm{2}+\sqrt{\mathrm{1}+\mathrm{2}{R}}=\sqrt{\mathrm{3}}{R} \\ $$$$\mathrm{1}+\mathrm{2}{R}=\mathrm{3}{R}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}\sqrt{\mathrm{3}}{R} \\ $$$$\mathrm{3}{R}^{\mathrm{2}} −{R}\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{2}\right)+\mathrm{3}=\mathrm{0} \\ $$$${R}=\frac{\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{2}\right)\pm\sqrt{\mathrm{48}+\mathrm{4}+\mathrm{16}\sqrt{\mathrm{3}}−\mathrm{36}}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}}\pm\frac{\mathrm{2}\sqrt{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}}{\mathrm{3}} \\ $$

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