Question

$\huge \bf Question$
A particle is moving around in a circle and its position is given in polar coordinates as x = Rcosθ, and y = Rsinθ, where R is the radius of the circle, and θ is in radians. From these equations derive the equation for centripetal acceleration.
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Answer 1

We only need to look at the equation for the x-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when θ = 0, the second derivative of x with respect to time must be the centripetal acceleration.

The first derivative of x with respect to time t is:

dx/dt = -Rsinθ(dθ/dt)

The second derivative of x with respect to time t is:

d^2x/dt^2 = -Rcosθ(dθ/dt)^2−Rsinθ(d^2θ/dt^2)

In both of the above equations the chain rule of Calculus is used and by assumption θ is a function of time. Therefore, θ can be differentiated with respect to time.

Now, evaluate the second derivative at θ = 0.

We have,

d^2x/dt^2 = -R(dθ/dt)^2

dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. we can set w ≡ dθ/dt.

Therefore,

d^2x/dt2 = -Rw^2

This is the well-known form for the centripetal acceleration equation.

@gurmanpreet1023

Answer 2

$\underline{\underline{\maltese\:\:\textbf{\textsf{Question}}}}$

• A particle is moving around in a circle and its position is given in polar coordinates as $\sf{x=R\cos\theta}$, and $\sf{y=R\sin\theta}$, where $\sf{R}$ is the radius of the circle, and $\sf{\theta}$ is in radians. From these equations derive the equation for centripetal acceleration

$\underline{\underline{\maltese\:\:\textbf{\textsf{Given}}}}$

• Polar Coordinate : $\sf{x=R\cos\theta}$
• Polar Coordinate : $\sf{y=R\sin\theta}$
• Where $\sf{R}$ is the radius of the circle, and $\sf{\theta}$ is in radians

$\underline{\underline{\maltese\:\:\textbf{\textsf{To Derive}}}}$

• The equation for centripetal acceleration from given equations

$\underline{\underline{\maltese\:\:\textbf{\textsf{Answer}}}}$

• Equation for centripetal acceleration from given equations : $\bf{ a=\omega^{2} R}$

$\underline{\underline{\maltese\:\:\textbf{\textsf{Calculations}}}}$

$\bigstar$ The position of a particle in circular motion in terms of polar coordinates is given by :

• $\sf{x=R\cos\theta\:\:------(1)}$
• $\sf{y=R\sin\theta\:\: ------(2)}$

➤ Here, $\sf{R}$ is the radius of the circle along which the particle moves and $\sf{\theta}$ is the angle made by the radius vector having magnitude $\sf{R}$ with x - axis

➠ If $\sf{t}$ is the time taken by the particle to move through angle $\sf{\theta}$, the angular velocity of the particle is, $\sf{\omega=\theta/t}$ ; Hence, we can write it as following

• $\sf{\theta=\omega t \:\: ------(3)}$

➠ Now, the position coordinates of the particle can be expressed as :

• $\sf{x=R\cos\omega t\:\: ------(4)}$
• $\sf{y=R\sin\omega t\:\:------(5)}$

$\underline{\normal{\textsf{ \bigstar\:\sf{D}\mathfrak{ifferentiate\: the\: above \: equations \: to\: find \: the\: components\: of \: velocit}\sf{y} }}}$

➠ Differentiating Equation 4 with respect to time t, the x-component of velocity $\sf{V_x}$ can be expressed as ; $\sf{\frac{d x}{d t}=\frac{d}{d t}(R \cos \omega t)}$

• $\sf{V_{x}=-\omega R \sin \omega t\:\:------(6)}$

➠ Differentiating Equation 5 with respect to time t, the y-component of velocity $\sf{V_y}$ can be expressed as ; $\sf{\frac{d y}{d t}=\frac{d}{d t}(R \sin \omega t)}$

• $\sf{V_{y}=\omega R \cos \omega t\:\:------(7)}$

➤ We know that acceleration is the change in velocity with time, differentiate Equation 6 with respect to time t to obtain the x-component of acceleration $\sf{a_x}$ ; $\sf{\frac{d V_{x}}{d t}=-\omega R \sin \omega t}$ $\implies\sf{a_{x}=-\omega^{2} R \cos \omega t}$

• $\sf{a_{x}=-\omega^{2} x \:\:-----(8)}$

➠ Also, differentiate Equation 7 with respect to time t to obtain the y -component of acceleration $\sf{a_y}$ ; $\sf{\frac{d V_{y}}{d t}=-\omega^{2} R \sin \omega t}$ $\implies\sf{a_{y}=-\omega^{2} R \sin \omega t}$

• $\sf{a_{y}=-\omega^{2} y \:\:------(9)}$

$\bigstar$ Using Equations 8 and 9, the magnitude of centripetal acceleration can be found as, (The Process is as follows)

$\sf{a=\sqrt{a_{x}\:^{2}+a_{y}\:^{2}}}$

$\to\sf{a=\sqrt{\left(-\omega^{2} x\right)^{2}+\left(-\omega^{2} y\right)^{2}}}$

$\to\sf{a=\sqrt{\omega^{4} x^{2}+\omega^{4} y^{2}}}$

$\to\sf{a=\sqrt{\omega^{4}\left(x^{2}+y^{2}\right)}}$

$\to\sf{a=\omega^{2} \sqrt{\left(x^{2}+y^{2}\right)}}$

➤ Since $\sf{x=R\cos\theta}$ and $\sf{y=R\sin\theta}$, We can write it as ;

$\to\sf{a=\omega^{2}\sqrt{R^{2} \cos ^{2} \theta+R^{2} \sin ^{2} \theta}}$

$\to\sf{a=\omega^{2}\sqrt{R^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}}$

$\to\sf{a=\omega^{2}\sqrt{R^{2}\left(1\right)}}$

$\to\sf{a=\omega^{2}\sqrt{R^{2}}}$

$\boxed{\boxed{\bf{\to a=\omega^{2} R}}}$

• Since the x and y components of acceleration are negative, it is evident that the centripetal acceleration is along the negative radial direction. Therefore, the centripetal acceleration can be expressed as $\sf{a=-\omega^{2} R}$ (The negative sign implies that the acceleration is directed towards the center of the circle)