1.18 Velocity (application)  (Page 5/2)

Problem : A particle is executing motion along a circle of radius “a” with a constant angular speed “ω” as shown in the figure. If the particle is at “O” at t = 0, then determine the position vector of the particle at an instant in xy - plane with "O" as the origin of the coordinate system.

Solution : Let the particle be at position “P” at a given time “t”. Then the position vector of the particle is :

$$\begin{array}{l}\mathbf{r}=x\mathbf{i}+y\mathbf{j}\end{array}$$

Note that "x" and "y" components of position vector is measured from the origin "O". From the figure,

$$\begin{array}{l}y=a\sin \theta =a\sin \omega t\end{array}$$

and

$$\begin{array}{l}x=a-a\cos \omega t=a(1-\cos \omega t)\end{array}$$

Thus, position vector of the particle in circular motion is :

$$\begin{array}{l}\mathbf{r}=a(1-\cos \omega t)\mathbf{i}+a\sin \omega t\mathbf{j}\end{array}$$

Problem : The position vector of a particle (in meters) is given as a function of time as :

$$\begin{array}{l}\mathbf{r}=2t\mathbf{i}+2t^2\mathbf{j}\end{array}$$

Determine the time rate of change of the angle “θ” made by the velocity vector with positive x-axis at time, t = 2 s.

Solution : It is a two dimensional motion. The figure below shows how velocity vector makes an angle "θ" with x-axis of the coordinate system. In order to find the time rate of change of this angle "θ", we need to express trigonometric ratio of the angle in terms of the components of velocity vector. From the figure :

$$\begin{array}{l}\tan \theta =\frac{v_y}{v_x}\end{array}$$

As given by the expression of position vector, its component in coordinate directions are :

$$\begin{array}{l}x=2t\mathrm{and}y=2t^2\end{array}$$

We obtain expression of the components of velocity in two directions by differentiating "x" and "y" components of position vector with respect to time :

$$\begin{array}{l}{v}_x=2\mathrm{and}{v}_y=4t\end{array}$$

Putting in the trigonometric function, we have :

$$\begin{array}{l}\tan \theta =\frac{v_y}{v_x}=\frac{4t}{2}=2t\end{array}$$

Since we are required to know the time rate of the angle, we differentiate the above trigonometric ratio with respect to time as,

$$\begin{array}{l}{\sec }^2\theta \frac{d\theta }{dt}=2\end{array}$$

$$\begin{array}{l}\Rightarrow (1+{\tan }^2\theta )\frac{d\theta }{dt}=2\\ \Rightarrow (1+4t^2)\frac{d\theta }{dt}=2\\ \Rightarrow \frac{d\theta }{dt}=\frac{2}{(1+4t^2)}\end{array}$$

At t = 2 s,

$$\begin{array}{l}\Rightarrow \frac{d\theta }{dt}=\frac{2}{(1+4x{2}^2)}=\frac{2}{17}\mathrm{rad}/s\end{array}$$

Problem : The displacement (x) of a particle is given by :

$$\begin{array}{l}x=A\sin (\omega t+\theta )\end{array}$$

At what time is the displacement maximum?

Solution : The displacement (x) depends on the value of sine function. It will be maximum for maximum value of sin(wt + θ). The maximum value of sine function is 1. Hence,

$$\begin{array}{l}\sin (\omega t+\theta )=1=\sin \left(\frac{\pi }{2}\right)\\ \Rightarrow \omega t+\theta =\pi 2\\ \Rightarrow t=\frac{\pi }{2\omega }-\frac{\theta }{\omega }\end{array}$$