Question
At any instant, the coordinates of a particle moving with constant angular speed in a circular path are $x= A \cos \omega t$ and $y= A \sin \omega t$. What is the radius of circular path?
✓
Answer
It is given,
$\begin{array}{l}
x=A \cos \omega t .....(1)\\
y=A \sin \omega t......(2)
\end{array}$
From equations (1) and (2)
$\begin{array}{cc}
& x^2+y^2=A^2 \cos ^2 \omega t+A^2 \sin ^2 \omega t \\
\Rightarrow \quad & x^2+y^2=A^2\left(\cos ^2 \omega t+\sin ^2 \omega t\right) \\
\Rightarrow \quad & x^2+y^2=A^2 \times 1 \\
& \because \sin ^2 \omega t+\cos ^2 \omega t=1 \\
\Rightarrow \quad & x^2+y^2=A^2.....(3)
\end{array}$
Equation (3) is equation of circle. Hence the centre of this circle is $(0,0)$ and radius of circular path is A .
$\begin{array}{l}
x=A \cos \omega t .....(1)\\
y=A \sin \omega t......(2)
\end{array}$
From equations (1) and (2)
$\begin{array}{cc}
& x^2+y^2=A^2 \cos ^2 \omega t+A^2 \sin ^2 \omega t \\
\Rightarrow \quad & x^2+y^2=A^2\left(\cos ^2 \omega t+\sin ^2 \omega t\right) \\
\Rightarrow \quad & x^2+y^2=A^2 \times 1 \\
& \because \sin ^2 \omega t+\cos ^2 \omega t=1 \\
\Rightarrow \quad & x^2+y^2=A^2.....(3)
\end{array}$
Equation (3) is equation of circle. Hence the centre of this circle is $(0,0)$ and radius of circular path is A .
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