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?

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Talia Downs

3 July, 23:51

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?

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Danielle Mcintosh · Answer 1

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