Question

A rigid body is rotating with a constant angular speed of 3 rad.s$^{-1}$ about a fixed axis passing through the points $A$ and $B$ with coordinates $(0, -1, 1)$ and $(1, 1, 3)$ respectively. Assuming all quantities in S.I. units, the instantaneous speed of the point $P$ of the body with coordinates $(4, 6, 7)$ is

• A. 3
• B. 2
• C. 6
• D. 4

Answer 1

Answer: (A)

3

Answer 2

The instantaneous speed of a point on a rigid body rotating about a fixed axis is given by the formula $v = \omega r$, where $\omega$ is the angular speed and $r$ is the perpendicular distance of the point from the axis of rotation.

1. Determine the axis of rotation:

The axis passes through points $A(0, -1, 1)$ and $B(1, 1, 3)$. The direction vector of the axis, $\vec{L}$, can be found by calculating $\vec{AB}$: $\vec{L} = \vec{AB} = B - A = (1-0, 1-(-1), 3-1) = (1, 2, 2)$.

2. Calculate the magnitude of the axis direction vector:

$|\vec{L}| = |\vec{AB}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

3. Determine the position vector of point P relative to a point on the axis:

Let's use point A as the reference. The vector $\vec{AP}$ is: $\vec{AP} = P - A = (4-0, 6-(-1), 7-1) = (4, 7, 6)$.

4. Calculate the perpendicular distance (r) of point P from the axis:

The perpendicular distance $r$ is the magnitude of the component of $\vec{AP}$ that is perpendicular to $\vec{AB}$. This can be found using the formula for the distance of a point from a line in 3D space: $r = \frac{|\vec{AP} \times \vec{AB}|}{|\vec{AB}|}$

First, calculate the cross product $\vec{AP} \times \vec{AB}$: $\vec{AP} \times \vec{AB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 7 & 6 \\ 1 & 2 & 2 \end{vmatrix}$ $= \mathbf{i}(7 \times 2 - 6 \times 2) - \mathbf{j}(4 \times 2 - 6 \times 1) + \mathbf{k}(4 \times 2 - 7 \times 1)$ $= \mathbf{i}(14 - 12) - \mathbf{j}(8 - 6) + \mathbf{k}(8 - 7)$ $= 2\mathbf{i} - 2\mathbf{j} + 1\mathbf{k} = (2, -2, 1)$

Next, calculate the magnitude of the cross product: $|\vec{AP} \times \vec{AB}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

Now, calculate the perpendicular distance $r$: $r = \frac{3}{3} = 1 \, \text{m}$.

5. Calculate the instantaneous speed of point P:

Given angular speed $\omega = 3 \, \text{rad.s}^{-1}$. The instantaneous speed $v = \omega r = (3 \, \text{rad.s}^{-1}) \times (1 \, \text{m}) = 3 \, \text{m.s}^{-1}$.