This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
To find the velocity of a particle in a rigid body spinning about an axis, we use the formula v = × r, where is the angular velocity vector and r is the position vector from a point on the axis of rotation to the particle's position. Step 1: Determine the angular velocity vector . The magnitude of the angular velocity is given as || = 3 rad/s. The axis of rotation is parallel to the vector u = 2i - 3k. First, find the magnitude of u: |u| = sqrt(2^2 + (-3)^2) = sqrt(4 + 9) = sqrt(13) The unit vector in the direction of the axis is u = u|u| = 2i - 3ksqrt(13). Now, form the angular velocity vector: = || u = 3 ( 2i - 3ksqrt(13) ) = (6)/(sqrt(13))i - (9)/(sqrt(13))k Step 2: Determine the position vector r. The axis passes through point A = (1, -3, -1). The particle is at point P = (3, -4, 1). The position vector r from a point on the axis (A) to the particle (P) is r = P - A. r = (3-1)i + (-4 - (-3))j + (1 - (-1))k r = 2i + (-4+3)j + (1+1)k r = 2i - j + 2k Step 3: Calculate the velocity v = × r. v = ( (6)/(sqrt(13))i - (9)/(sqrt(13))k ) × (2i - j + 2k) Factor out (1)/(sqrt(13)): v = (1)/(sqrt(13)) (6i + 0j - 9k) × (2i - j + 2k) Compute the cross product: (6i - 9k) × (2i - j + 2k) = i & j & k \\ 6 & 0 & -9 \\ 2 & -1 & 2 = ((0)(2) - (-9)(-1))i - ((6)(2) - (-9)(2))j + ((6)(-1) - (0)(2))k = (0 - 9)i - (12 - (-18))j + (-6 - 0)k = -9i - (12 + 18)j - 6k = -9i - 30j - 6k Now, multiply by the scalar factor: v = (1)/(sqrt(13)) (-9i - 30j - 6k) v = -(9)/(sqrt(13))i - (30)/(sqrt(13))j - (6)/(sqrt(13))k m/s The velocity of the particle is -(9)/(sqrt(13))i - (30)/(sqrt(13))j - (6)/(sqrt(13))k m/s. Drop the next question.