# Linear velocity of a rotating body

1. Jul 30, 2009

### rugapark

1. The problem statement, all variables and given/known data

A flat rigid body is rotating with angular velocity 3 rads-1 about an axis in the
direction of the vector (i + 2 j + 3 k) and passing through the point (1, 1, 0) on
the body. Find the linear velocity of the point P = (1, 0, 1) on the body.
(You may use the result v = w x r .)

2. Relevant equations

v= w x r

3. The attempt at a solution

i have no idea where to go with this - i need to find r, but not sure how to go about using the coordinates given.

2. Jul 30, 2009

### HallsofIvy

Staff Emeritus
Do you mean 3 rads/sec (often written just "3 s-1") ?

First you need to know the radius of the circle the point is moving in. Draw a line from (1, 0, 1) to the line x= 1+ t, y= 1+ 2t, z= 3t. The plane containing (1, 0, 1) and perpendicular to i+ 2j+ 3k is (x-1)+ 2y+ 3(z-1)= 0. The line passes through that plane at (1+ t- 1)+ 2(1+ 2t)+ 3(3t-1)= 14t- 1= 0 or t= 1/14. x= 1+ 1/14, y= 1+ 2/14, z= 3/14 or (15/14, 16/14, 3/14). The distance from that point to (1, 0, 1) is
$$\sqrt{(1- 15/14)^2+ (-16/14)^2+ (1- 3/14)^2}$$
$$= \sqrt{1/196+ 256/196+ 121/196}$$
$$= 3\sqrt{42}/14$$
and that is the radius of the circle the point is moving in. (Better check my arithmetic- that looks peculiar.) From the radius you can calculate the distance corresponding to 3 radians and so the distance the particle moves in one second.

3. Aug 1, 2009

### rugapark

how did you get the x, y and z to equal those three? and where did the t's come from?

thanks for the help!