Kinematics 6 Question 4
4. In three dimensional system, the position coordinates of a particle (in motion) are given below
$$ \begin{aligned} x & =a \cos \omega t \\ y & =a \sin \omega t \\ z & =a \omega t \end{aligned} $$
The velocity of particle will be
(a) $\sqrt{2} a \omega$
(b) $2 a w$
(c) $a \omega$
(d) $\sqrt{3} a \omega$
(2019 Main, 09 Jan II)
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Answer:
Correct Answer: 4. (a)
Solution:
- Given that the position coordinates of a particle
$$ \begin{aligned} x & =a \cos \omega t \\ y & =a \sin \omega t \\ z & =a \omega t \end{aligned} $$
So, the position vector of the particle is
$$ \Rightarrow \quad \begin{aligned} \hat{\mathbf{r}} & =x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \hat{\mathbf{r}} & =a \cos \omega t \hat{\mathbf{i}}+a \sin \omega t \hat{\mathbf{j}}+a \omega t \hat{\mathbf{k}} \\ \hat{\mathbf{r}} & =a[\cos \omega t \hat{\mathbf{i}}+\sin \omega t \hat{\mathbf{j}}+\omega t \hat{\mathbf{k}}] \end{aligned} $$
therefore, the velocity of the particle is
$$ \begin{array}{rlrl} \because & & \hat{\mathbf{v}} & =\frac{d \mathbf{r}}{d t}=\frac{d[a][\cos \omega t \hat{\mathbf{i}}+\sin \omega t \hat{\mathbf{j}}+\omega t \hat{\mathbf{k}}]}{d t} \\ \Rightarrow & \hat{\mathbf{v}} & =-a \omega \sin \omega t \hat{\mathbf{i}}+a \omega \cos \omega t \hat{\mathbf{j}}+a \omega \hat{\mathbf{k}}) \end{array} $$
The magnitude of velocity is
$$ \begin{aligned} |\mathbf{v}| & =\sqrt{v _x^{2}+v _y^{2}+v _z^{2}} \\ \text { or } \quad|\mathbf{v}| & =\sqrt{(-a \omega \sin \omega t)^{2}+(a \omega \cos \omega t)^{2}+(a \omega)^{2}} \\ & =\omega a \sqrt{(-\sin \omega t)^{2}+(\cos \omega t)^{2}+(1)^{2}}=\sqrt{2} \omega a \end{aligned} $$
