same observers O and B as in the previous example, we have R-2m92x(vA/B)xyx-m92×(9×TA/B)=m(aA/B)xy Now,SX(UA/B)xy2=-Qver, nx( Q xrA/B)=-2rer, and(aA/B)xy2=-(u2/r)er, thus, R=-(2m9v+9r+-)er The term(v2/r)is the acceleration of A with respect to the table, the term 222r is the acceleration of a point situated at A and fixed with respect to the table, and, finally, the term 2Q u is extra. This final term is the acceleration due to the Coriolis effect. It is present because B is rotating, and it is required to obtain the “ correct” result for acceleration.same observers O and B as in the previous example, we have R − 2mΩ × (vA/B)x′y′z ′ − mΩ × (Ω × rA/B) = m(aA/B)x′y′z ′ . Now, Ω × (vA/B)x′y′z ′ = −Ωv er, Ω × (Ω × rA/B) = −Ω 2 r er, and (aA/B)x′y′z ′ = −(v 2/r) er, thus, R = −(2mΩv + Ω2 r + v 2 r ) er = − (Ωr + v) 2 r er. The term (v 2/r) is the acceleration of A with respect to the table, the term Ω2 r is the acceleration of a point situated at A and fixed with respect to the table, and, finally, the term 2Ωv is extra. This final term is the acceleration due to the Coriolis effect. It is present because B is rotating, and it is required to obtain the “correct” result for acceleration. 7