Question

The linear velocity of a rotating body is given by $\vec{v} = \vec{\omega} \times \vec{r}$, where $\vec{\omega}$ is the angular velocity and $\vec{r}$ is radius vector. The angular velocity of a body is $\vec{\omega} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ and the radius vector $\vec{r} = 4 \hat{j} - 3 \hat{k}$, then $| \vec{v} |$ is:
A. $\sqrt{29} units$
B. $\sqrt{31} units$
C. $\sqrt{37} units$
D. $\sqrt{41} units$

Answer 1

Hint: We are given a radius vector and angular velocity vector, we simply have to find its cross product like in the formula mentioned in the question. Then we must find the magnitude of the obtained cross product vector.

Complete answer:
If we are given two vectors such that:
$\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$
$\vec{q} = d \hat{i} + e \hat{j} + f \hat{k}$

Then their cross product is given as:
$\vec{p} \times \vec{q} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
a & b & c\\
d & e & f
\end{vmatrix}
$

Now, in analogy to this example, we can use $\vec{\omega}$ in place of $\vec{p}$ and $\vec{r}$ in place of $\vec{q}$ .
So, we are basically given that:
a = 1, b = -2, c = 2;
d = 0, e = 4, f = -3.
Substituting these values we get:
$\vec{\omega} \times \vec{r} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
1 & -2 & 2\\
0 & 4 & -3
\end{vmatrix}

= \hat{i} (6 -8) - \hat{j} (-3 -0) + \hat{k}(4 -0)
$
So we can write the linear velocity vector as:
$\vec{v} = -2 \hat{i} +3 \hat{j} + 4 \hat{k}$
The magnitude of any vector can be obtained by taking a dot product of the vector with itself which will give us a number (positive). If we take the square root of this dot product, we get the magnitude of the vector.
Therefore, the magnitude of the obtained velocity vector can be written as:
$ | \vec{v} | = \sqrt{\vec{v} . \vec{v}} = \sqrt{(-2).(-2) + (3).(3) + (4).(4)} = \sqrt{4 + 9 + 16} = \sqrt{29}$

So, the correct answer is “Option A”.

Additional Information:
The cross product of any two vectors gives a resultant vector which has a direction perpendicular to both the vectors. In our case we were given an angular velocity and radius vector so our velocity vector will be in a direction perpendicular to both these vectors. Radius vector is in the direction of radius of a circle and angular velocity is in the plane perpendicular to the circle. Therefore we get the direction of linear velocity as tangential to the radius vector in case of a circular motion. A dot product always results in a scalar quantity.

Note:
While taking the cross product, the placement for various vector vectors along x, y and z directions should be properly noted. Like in the case of the radius vector, we could see that x component was missing, so we put a zero in its place. One might easily get a wrong answer if one would have made a simple error in noting this down properly. It is suggested that one should at least check the values twice.