Question

Q. find the radial and transverse components of a particle moving along the circle
square of x plus square of y is equal to square of a.

Answer 1

Answer on Question #50225, Physics, Mechanics | Kinematics | Dynamics

Find the radial and transverse components of acceleration of a particle moving along the circle $x^{2} + y^{2} = a^{2}$ with a constant velocity $c$.

Solution:

Given that

$$\frac {d \theta}{d t} = c$$

Differentiate w.r.t "t" we get

$$\frac {d ^ {2} \theta}{d t ^ {2}} = 0$$

Also given that

$$x ^ {2} + y ^ {2} = a ^ {2}$$

First we change this into polar form by putting $x = r\cos \theta$ and $y = r\sin \theta$

$$\begin{array}{l} r ^ {2} \cos^ {2} \theta + r ^ {2} \sin^ {2} \theta = a ^ {2} \\ \Rightarrow r ^ {2} \left(\cos^ {2} \theta + \sin^ {2} \theta\right) = a ^ {2} \\ \Rightarrow r ^ {2} = a ^ {2} \\ \Rightarrow r = a \\ \Rightarrow \frac {d r}{d t} = 0 \quad = 2 \frac {d ^ {2} r}{d t ^ {2}} = 0 \\ \end{array}$$

Radial component of acceleration $= a_{r} = \frac{d^{2}r}{dt^{2}} - r\left(\frac{d\theta}{dt}\right)^{2} = 0 - ac^{2} = -ac^{2}$

Transverse component of acceleration $= a_{\theta} = 2\frac{dr}{dt}\left(\frac{d\theta}{dt}\right) + r\frac{d^2\theta}{dt^2} = 0$

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