Physics - Center of Mass Question with Solution | TestHub

Initially, the particle is at the lowest point, so its velocity relative to the ground is zero. The initial kinetic energy of the system (sphere + particle) is $K{E}_i=\frac{1}{2}M{V}_0^2+\frac{1}{2}I{\omega }_0^2=\frac{1}{2}M(R{\omega }_0{)}^2+\frac{1}{2}(\frac{2}{3}M{R}^2){\omega }_0^2=\frac{5}{6}M{R}^2{\omega }_0^2$. Taking the horizontal surface as reference, initial potential energy $P{E}_i=MgR$. At maximum height, the particle is at the top of the sphere, and for maximum height, the system momentarily comes to rest, so final kinetic energy $K{E}_f=0$. If the sphere's center of mass rises by $H$, its final potential energy is $Mg(R+H)$, and the particle's final potential energy is $mg(2R+H)$ (since its height is $2R+H$). By conservation of energy, $K{E}_i+P{E}_i=K{E}_f+P{E}_f⟹\frac{5}{6}M{R}^2{\omega }_0^2+MgR=0+Mg(R+H)+mg(2R+H)$. This simplifies to $\frac{5}{6}M{R}^2{\omega }_0^2=(M+m)gH+2mgR$. Solving for $H$, we get $H=\frac{\frac{5}{6}M{R}^2{\omega }_0^2-2mgR}{(M+m)g}$. The maximum height attained by the particle from its initial position (ground) is $h_{max}=2R+H=2R+\frac{\frac{5}{6}M{R}^2{\omega }_0^2-2mgR}{(M+m)g}=\frac{2R(M+m)g+\frac{5}{6}M{R}^2{\omega }_0^2-2mgR}{(M+m)g}=\frac{2MgR+\frac{5}{6}M{R}^2{\omega }_0^2}{(M+m)g}=\frac{MR(2g+\frac{5}{6}R{\omega }_0^2)}{(M+m)g}$.