How fast, in rpm, would a 4.0 kg , 27-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s ?

Well, we know that angular momentum is L= I*w, where I=(moment of inertia) and w=angular frequency. The moment of inertia of a spherical shell is 2/3*M*R^2. So we know that I=(2/3)(4kg)*(.135^2). we'll set L=0.26 kg*m^2/s=0.0486*w and solve for w. so w= 5.34979423868 m/s for rpm w=5.34979423868*60/(2*pi) so w= 51.0867718566 rpm