Rotation of an object using cross product

Question

Given a rigid body that is rotating about a fixed axis with constant angular speed ω, the angular velocity vector ω is defined as the vector with magnitude ω whose direction is parallel to the axis of rotation Then, if we assume that vector r represents the position of the object, we can calculate the linear speed of the object as

vector v = vector ω x vector r

and

magnitude of vector v = magnitude of (vector ω x vector r)

Now assume we have 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 < 2,1,2 >. 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 measured in meters.

Hint: Your final answer will contain ω.

Answer 1

(1) We are given a problem of rotational motion. The rotating object spins with constant speed ω around a fixed axe (say, axis z on the x-y-z coordinates.

Given----> ω : magnitude of constant angular speed

(2) The direction of the vector of ω passes through the point (2,1,2), because it starts from origin, stretches to that point. The unit vector of that direction in obtained by dividing the three coordinate lengths, 2,1,3, by the absolute distance from origin to that point. as follows:

vector of ω = ( 2 i + 1 j + 2 k) / sqrt (2^2 +1^2 + 2^2)

Given-----> vector of ω = ( 2 i + 1 j + 2 k) / 3)

Where i, j, and k are the unit vectors of x, y, and k, respectively.

(3) The position vector through the object is given by:

Given-----> vector r= 1 i + 3 j + 5 k

(4) The velocity v is then determined from the above three given facts as follows:

vector v = vector ω (cross product) vector r

Since the magnitude of ω is given as a variable, we could state the cross product as follows:

vector v = (ω / 3 ) x | i j k |

| 2 1 2 |

| 1 3 5 |

= (ω / 3 ) x ( i (5 - 6) - j (10 - 2) + k ( 6 - 1) )

= (ω / 3 ) x ( - i - 8 j + 5 k )

This is the answer in vector format:

vector v = = (ω / 3 ) x ( - i - 8 j + 5 k )

In scalar format, we square each term, add the three squares, then take their square root as follows

abs (v ) = (ω / 3 ) x sqrt ( 1^2 + 8^2 + 5^2 )

= (ω / 3 ) x sqrt ( 1 + 64 + 25 )

= (ω / 3 ) x sqrt ( 90 )

= ω x sqrt ( 10 ) ==================final answer =============