6.4 Accelerated motion in two dimensions  (Page 4/6)

Problem : The coordinates of a particle moving in a plane are given by x = A cos(ωt) and y = B sin (ωt) where A, B (<A) and ω are positive constants. Find the nature of path of motion.

Solution : We shall use the general technique to find path of motion in two dimensional case. In order to find the path motion, we need to have an equation that connects “x” and “y” coordinates of the planar coordinate system. Note that there is no third coordinate.

Elliptical motion

An inspection of the expressions of “x” and “y” suggests that we can use the trigonometric identity,

$$\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\end{array}$$

Here, we have :

$$\begin{array}{l}x=A\cos \left(\omega t\right)\\ \Rightarrow \cos \left(\omega t\right)=\frac{x}{A}\end{array}$$

Similarly, we have :

$$\begin{array}{l}y=B\sin \left(\omega t\right)\\ \Rightarrow \sin \left(\omega t\right)=\frac{y}{B}\end{array}$$

Squaring and adding two equations,

$$\begin{array}{l}{\sin }^2\left(\omega t\right)+{\cos }^2\left(\omega t\right)=1\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{x^2}{A^2}+\frac{y^2}{B^2}=1\end{array}$$

This is an equation of ellipse. Hence, the particle follows an elliptical path.

Problem : The coordinates of a particle moving in a plane are given by x = A cos(ωt) and y = B sin (ωt) where A, B (<A) and ω are positive constants. Investigate the nature of velocity and acceleration for this motion. Also, discuss the case for A = B and when "ω" is constant.

Solution : We can investigate the motion as required if we know expressions of velocity and acceleration. Therefore, we need to determine velocity and acceleration. Since components of position are given, we can find components of velocity and acceleration by differentiating the expression with respect to time.

1: Velocity

The components of velocity in “x” and “y” directions are :

$$\begin{array}{l}\frac{đx}{đt}=v_x=-A\omega \sin \left(\omega t\right)\\ \frac{đy}{đt}=v_y=B\omega \cos \left(\omega t\right)\end{array}$$

The velocity of the particle is given by :

$$\begin{array}{l}\Rightarrow \mathbf{v}=\omega \{-A\sin (\omega t)\mathbf{i}+B\cos (\omega t\left)\mathbf{j}\right\}\end{array}$$

Evidently, magnitude and direction of the particle varies with time.

2: Acceleration

We find the components of acceleration by differentiating again, as :

$$\begin{array}{l}\frac{{đ}^{2}x}{đt^2}=a_x=-A{\omega }^2\cos \left(\omega t\right)\\ \frac{{đ}^{2}y}{đt^2}=a_y=-B{\omega }^2\sin \left(\omega t\right)\end{array}$$

Both “x” and “y” components of the acceleration are trigonometric functions. This means that acceleration varies in component direction. The net or resultant acceleration is :

$$\begin{array}{l}\Rightarrow \mathbf{a}=-{\omega }^2\left\{A\cos \right(\omega t)\mathbf{i}+B\sin (\omega t\left)\mathbf{j}\right\}\end{array}$$

3: When A = B and "ω" is constant

When A = B, the elliptical motion reduces to circular motion. Its path is given by the equation :

$$\begin{array}{l}\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{x^2}{A^2}+\frac{y^2}{A^2}=1\end{array}$$

$$\begin{array}{l}\Rightarrow x^2+y^2=A^2\end{array}$$

This is an equation of circle of radius “A”. The speed for this condition is given by :

$$\begin{array}{l}v=\surd \left\{A^2{\omega }^2{\sin }^2\right(\omega t)+A^2{\omega }^2{\cos }^2(\omega t\left)\right\}\\ v=A\omega \end{array}$$

Thus, speed becomes a constant for circular motion, when ω = constant.

The magnitude of acceleration is :

$$\begin{array}{l}a={\omega }^2\surd \left\{A^2{\sin }^2\right(\omega t)+A^2{\cos }^2(\omega t\left)\right\}\\ a=A{\omega }^2\end{array}$$

Thus, acceleration becomes a constant for circular motion, when ω = constant.