A rigid body rotates with a constant angular speed of ω radians per second about a line through the origin in the direction of the vector 2i+j+2k. Find the speed of an object on this body at the instant the object passes through the point (1,3,5). Assume the distance is in meters.
Solution The vector ω is parallel to the axis of rotation of the rigid body, and so it is in the direction of the vector 2i+j+2k. Since 2i+j+2k√22+12+22=23i+13j+23k is a unit vector in the direction of 2i+j+2k, the angular velocity ω is ω=ω(23i+13j+23k)
The position of the object is r=i+3j+5k. So, the velocity v of the object is v=ω×r=[ω(23i+13j+23k)]×[i+3j+5k]=ω|ijk231323135|=ω(−13i−83j+53k)
The speed v of the object is v=‖ meters per second.