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(C) Since the sphere is rolling without slipping and there is no friction, the total mechanical energy is conserved.Let $R$ be the radius of the track

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{3}{10}\right)$

Vedclass · Answer

(C) Since the sphere is rolling without slipping and there is no friction, the total mechanical energy is conserved.Let $R$ be the radius of the track and $r$ be the radius of the sphere. Given $r \ll R$.The total energy at the lowest point (bottom) is the sum of translational and rotational kinetic energy:$E_{bottom} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere, $I = \frac{2}{5}mr^2$. Since it rolls without slipping, $\omega = \frac{v}{r}$.$E_{bottom} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2ight)\left(\frac{v}{r}ight)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$At the maximum height $h$ reached by the center of mass, the velocity becomes zero, so the total energy is purely potential energy:$E_{top} = mgh$Equating the energies: $\frac{7}{10}mv^2 = mgh \Rightarrow h = \frac{7v^2}{10g}$From the geometry of the track, $h = R(1 - \cos	heta)$.Substituting the values $v = 10 \ m/s$, $g = 10 \ m/s^2$, and $R = 10 \ m$:$h = \frac{7 	imes (10)^2}{10 	imes 10} = 7 \ m$$7 = 10(1 - \cos	heta) \Rightarrow 0.7 = 1 - \cos	heta \Rightarrow \cos	heta = 0.3 = \frac{3}{10}$Therefore, $	heta = \cos^{-1}\left(\frac{3}{10}ight)$.

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