Answer:
[tex]v=1.42\frac{m}{s}[/tex]
Explanation:
There is no friction in the physical system. Thus, according to the law of conservation of energy, recall that the object is released from rest:
[tex]\Delta E=0\\U=K_R+K_T\\mgh=\frac{I_p\omega_p^2}{2}+\frac{I_s\omega_s^2}{2}+\frac{mv^2}{2}[/tex]
Recall that the moment of inertia of a sphere is [tex]I_s=\frac{2MR^2}{3}[/tex]. The angular speed of the pulley is [tex]\omega_p=\frac{v}{r}[/tex] and the angular speed of the sphere is [tex]\omega_s=\frac{v}{R}[/tex]. So, we replace:
[tex]mgh=\frac{I_pv^2}{2r^2}+\frac{2MR^2}{3}\frac{v^2}{2R^2}+\frac{mv^2}{2}\\mgh=v^2(\frac{I_p}{2r^2}+\frac{M}{3}+\frac{m}{2})\\v^2=\frac{mgh}{\frac{I_p}{2r^2}+\frac{M}{3}+\frac{m}{2}}\\\\v=\sqrt{\frac{(0.6kg)(9.8\frac{m}{s^2})(0.82m)}{\frac{3*10^{-3}kg\cdot m^2}{2(0.05m)^2}+\frac{4.5kg}{3}+\frac{0.6kg}{2}}}\\v=1.42\frac{m}{s}[/tex]
