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(B) Let $m$ be the mass of the particle,$r$ be the radius of the circular path,$\omega$ be the angular velocity,and $	heta = 45^\circ$ be the semi-ve

Vedclass · Accepted Answer

$\sqrt{30} \ 	ext{rad/s}$

Vedclass · Answer

(B) Let $m$ be the mass of the particle,$r$ be the radius of the circular path,$\omega$ be the angular velocity,and $	heta = 45^\circ$ be the semi-vertical angle. The height $h = 1 \ m$. Since $	an 	heta = r/h$,we have $r = h 	an 45^\circ = 1 \ m$.For maximum angular velocity,the friction force $f = \mu N$ acts downwards along the slant surface to prevent the particle from sliding up.The forces acting on the particle are: Normal force $N$ perpendicular to the surface,gravitational force $mg$ downwards,and friction $f = \mu N$ downwards along the slope.Resolving forces horizontally: $N \sin 	heta + f \cos 	heta = m \omega^2 r$.Resolving forces vertically: $N \cos 	heta - f \sin 	heta = mg$.Substituting $f = \mu N$: $N(\sin 	heta + \mu \cos 	heta) = m \omega^2 r$ and $N(\cos 	heta - \mu \sin 	heta) = mg$.Dividing the two equations: $\frac{\sin 	heta + \mu \cos 	heta}{\cos 	heta - \mu \sin 	heta} = \frac{\omega^2 r}{g}$.Given $	heta = 45^\circ$,$\sin 45^\circ = \cos 45^\circ = 1/\sqrt{2}$,and $\mu = 0.5$:$\frac{1/\sqrt{2} + 0.5/\sqrt{2}}{1/\sqrt{2} - 0.5/\sqrt{2}} = \frac{\omega^2 (1)}{10} \implies \frac{1.5}{0.5} = \frac{\omega^2}{10} \implies 3 = \frac{\omega^2}{10}$.Thus,$\omega^2 = 30$,so $\omega = \sqrt{30} \ 	ext{rad/s}$.

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