Question Number 54462 by ajfour last updated on 03/Feb/19
Commented by ajfour last updated on 04/Feb/19
$${Find}\:{necessary}\:\omega_{\mathrm{0}} \:{that}\:{need}\:{be} \\ $$$${given}\:{to}\:{the}\:{sphere}\:{such}\:{that}\:{it} \\ $$$${reaches}\:{the}\:{vertical}\:{wall},\:{rolling}\: \\ $$$${purely}\:{along}\:{the}\:{incline}.\:\:\:\:\: \\ $$
Answered by mr W last updated on 04/Feb/19
![[b tan α+R tan ((π/4)−(α/2))−R]mg=(1/2)×(((2mR^2 )/5)+mR^2 )ω_0 ^2 [(b/R) tan α+tan ((π/4)−(α/2))−1]g=((7R)/(10))ω_0 ^2 ⇒ω_0 =(√(((10g)/(7R))[(b/R) tan α−((2tan (α/2))/(1+tan (α/2)))]))](https://www.tinkutara.com/question/Q54494.png)
$$\left[{b}\:\mathrm{tan}\:\alpha+{R}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right)−{R}\right]{mg}=\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{2}{mR}^{\mathrm{2}} }{\mathrm{5}}+{mR}^{\mathrm{2}} \right)\omega_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\left[\frac{{b}}{{R}}\:\mathrm{tan}\:\alpha+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right)−\mathrm{1}\right]{g}=\frac{\mathrm{7}{R}}{\mathrm{10}}\omega_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\Rightarrow\omega_{\mathrm{0}} =\sqrt{\frac{\mathrm{10}{g}}{\mathrm{7}{R}}\left[\frac{{b}}{{R}}\:\mathrm{tan}\:\alpha−\frac{\mathrm{2tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right]} \\ $$