a hollow sphere of mass 58.0 g and radius 11.8 cm rolls, starting from rest, without slipping down an…

Explanation:

Step1: Apply conservation of mechanical energy

Initial energy $E_i = mgh$ (potential energy only as starts from rest), where $m = 58.0\ g=0.058\ kg$, $g = 9.8\ m/s^2$, $h = 1.560\ m$. Final energy $E_f=K_{trans}+K_{rot}$. For a hollow - sphere, the moment of inertia $I=\frac{2}{3}mr^{2}$, and $v = r\omega$. So $K_{trans}=\frac{1}{2}mv^{2}$ and $K_{rot}=\frac{1}{2}I\omega^{2}=\frac{1}{2}(\frac{2}{3}mr^{2})\omega^{2}=\frac{1}{3}mr^{2}\omega^{2}$, and $v = r\omega$ so $K_{trans}=\frac{1}{2}mr^{2}\omega^{2}$. Then $E_f=\frac{1}{2}mr^{2}\omega^{2}+\frac{1}{3}mr^{2}\omega^{2}=\frac{3 + 2}{6}mr^{2}\omega^{2}=\frac{5}{6}mr^{2}\omega^{2}$. $mgh=\frac{5}{6}mr^{2}\omega^{2}$

Step2: Solve for $\omega$

Cancel out $m$ from both sides of the equation $mgh=\frac{5}{6}mr^{2}\omega^{2}$, we get $gh=\frac{5}{6}r^{2}\omega^{2}$. Then $\omega=\sqrt{\frac{6gh}{5r^{2}}}$. Given $r = 11.8\ cm = 0.118\ m$, $h = 1.560\ m$, $g = 9.8\ m/s^2$. $\omega=\sqrt{\frac{6\times9.8\times1.560}{5\times(0.118)^{2}}}$ $\omega=\sqrt{\frac{91.728}{5\times0.013924}}$ $\omega=\sqrt{\frac{91.728}{0.06962}}$ $\omega=\sqrt{1317.55}$ $\omega\approx36.3\ rad/s$

Answer:

36.3