# VECTORS

Q. If a rigid body is rotating about an axis passing through the point 2i^-j^-k^ and parallel to I^-2j^+2k^ with an angular velocity 3 radians/sec, then find the velocity of the point of the rigid body whose position vector is 2i^+3j^-4k^.
a) -2i^+3j^+4k^
b) 2i^-3j^+4k^
c) -2i^+3j^-4k^
d) -2i^-3j^-4k^

 jcaron2 Posts: 983, Reputation: 1034 Senior Member #2 May 9, 2011, 12:45 PM
Since this is a multiple choice question, we can do this the easy way! Since the specified point rotates AROUND the axis, we automatically know that its velocity vector is perpendicular to the axis. Hence, the dot product between the velocity vector and the vector parallel to the axis MUST BE ZERO.

a) $(-2i+3j+4k)\cdot(i-2j+2k)=-2-6+8=0$
b) $(2i-3j+4k)\cdot(i-2j+2k)=2+6+8=16$
c) $(-2i+3j-4k)\cdot(i-2j+2k)=-2-6-8=-16$
d) $(-2i-3j-4k)\cdot(i-2j+2k)=-2+6-8=-4$

 ApoorvGoel Posts: 36, Reputation: 10 Junior Member #3 May 19, 2011, 11:33 AM
Comment on jcaron2's post
Quote:
 Originally Posted by jcaron2 Since this is a multiple choice question, we can do this the easy way! Since the specified point rotates AROUND the axis, we automatically know that its velocity vector is perpendicular to the axis. Hence, the dot product between the velocity vector and the vector parallel to the axis MUST BE ZERO. a) $(-2i+3j+4k)\cdot(i-2j+2k)=-2-6+8=0$ b) $(2i-3j+4k)\cdot(i-2j+2k)=2+6+8=16$ c) $(-2i+3j-4k)\cdot(i-2j+2k)=-2-6-8=-16$ d) $(-2i-3j-4k)\cdot(i-2j+2k)=-2+6-8=-4$ Clearly the answer is (a).
Can you explain this question please
I am not able to understand it
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #4 May 19, 2011, 12:39 PM

The question says that you have a body under circular motion, okay?

Now, that body is rotating around an axis. The axis is parallel to the vector <1, -2, 2>, and hence has this very direction.

The axis is perpendicular to the direction of motion of the body, right? Jcaron used a method called dot product. This is something that can be used to check whether two directional vectors are perpendicular or not, giving a result of 0 when the vectors are perpendicular.

So what we need, is to find the velocity vector, which also gives the direction of the body. So, if we find the dot product of the two vectors (velocity and axis), we should get the result 0 since they are perpendicular. This is where the work that jcaron did comes into play, that is, finding the dot product of the two vectors, one being the directional vector of the velocity, <-2, 3, 4> and the others in parts b to d and the other being the direction of the axis vector, <1, -2, 2>.

I hope this makes sense.
 ebaines Posts: 10,585, Reputation: 5794 Expert #5 May 19, 2011, 12:52 PM

Jc showed a really easy way to do it. But suppose it wasn't a multiple choice question. We'd have to do it the hard way:

Use the relation:

$
\vec v = \vec \omega \ \times \ \vec r
$

where $\vec \omega$ is the vector in the direction of $(\hat i-2 \hat j+2 \hat k)$ that has magnitude of 3 radians. But it turns out that the magnitude of $(\hat i-2 \hat j+2 \hat k)$ is 3, so we're good to go. And $\vec r$ is the vector from any point along the spin axis to the point in question:

$
\vec r = (2 \hat i +3 \hat j - 4 \hat k) - ( 2 \hat i - \hat j - \hat k) = 4 \hat j -3 \hat k
$

So
$
\vec \omega = (\hat i-2 \hat j+2 \hat k) \ \times \ (4 \hat j -3 \hat k) = -2 \hat i + 3 \hat j + 4 \hat k
$

Which agrees with choice (a).

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