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A regid body sping with angular velocity of 4 radias per sec about axis parallel to (3j-k) passing through the point (I+3j+k) .find find the velocity of the partical at a point (4i-2j+k)
Obviously,the unit vector (^n) in the direction of 3j – k = (3j – k )/l 3j – k l =(3j – k)/sqrt(10)angular velocity of a rigid body= 4*[(3j – k )/sqrt(10)]= 12j – 4k / sqrt(10)Let the point whose velocity is desired to be determined be P.Then its +ve w.r.t. the point 2i-j-k on the axis given byr=(4i – 2j + k)-(i + 3j + k)= 3i – 5jHence,if v be the velocity of pt. P,we have v=w(omegha)cross r = (3j – k) cross (3i – 5j) = – 5i – 3j– 9k
Obviously,the unit vector (^n) in the direction of
3j – k = (3j – k )/l 3j – k l
=(3j – k)/sqrt(10)
angular velocity of a rigid body= 4*[(3j – k )/sqrt(10)]= 12j – 4k / sqrt(10)
Let the point whose velocity is desired to be determined be P.Then its +ve w.r.t. the point 2i-j-k on the axis given by
r=(4i – 2j + k)-(i + 3j + k)= 3i – 5j
Hence,if v be the velocity of pt. P,we have v=w(omegha)cross r
= (3j – k) cross (3i – 5j) = – 5i – 3j– 9k
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