Centripetal Force and Conical Pendulum

The acceleration in uniform circular motion is always directed towards the centre of the circle. The centripetal acceleration or radial acceleration represents the acceleration towards the center along the radius of the circle.

The centripetal acceleration has been already defined in circular motion. The centripetal acceleration of a particle of mass $m$ in uniform circular motion is ${{a}_{\text{cen}}}=\frac{{{v}^{2}}}{R}$ if $R$ is the radius of the circle. Therefore, the centripetal force is

\[{{F}_{\text{cen}}}=\frac{m{{v}^{2}}}{R} \tag{1} \label{1}\]

If the particle completes one rotation in the circle in time period $T$, it travels a distance equal to the circumference of the circle which is $2\pi R$. So, the speed of the particle is $v=\frac{2\pi R}{T}$. Now putting the value of $v=\frac{2\pi R}{T}$ in Eq. \eqref{1} we get,

\[{{F}_{\text{cen}}}=\frac{4m{{\pi }^{2}}R}{{{T}^{2}}} \tag{2} \label{2}\]

The above equation gives us the centripetal force or the radial force.

Conical Pendulum

Now it's time to talk some Physics about a conical pendulum. For this consider a pendulum of a bob of mass $m$. The bob is hanged by a massless thread of length $L$. Note that the massless thread is only an idealization. No thread is massless in the world. The bob moves along a horizontal circle of radius $R$ with constant speed $v$.

Figure 1 Conical pendulum

In Figure 1 the weight of the bob acts vertically downwards which is $\overrightarrow w = m\overrightarrow g $. The tension ${{T}_{F}}$ in the thread has two components. The vertical component of ${{T}_{F}}$ is ${{T}_{F}}\cos \alpha $ and the horizontal component is ${{T}_{F}}\sin \alpha $. The vertical component balances the weight and the horizontal component gives the centripetal force to the bob moving in the circle. Here we find the period of the pendulum and how it is related to the angle $\alpha$ the bob makes with the vertical line. You can get,

\[\begin{align} {{T}_{F}}\cos \alpha &=w \tag{3} \label{3}\\ \text{and,}\quad {{T}_{F}}\sin \alpha & =\frac{m{{v}^{2}}}{R} \tag{4} \label{4} \end{align}\]

If ${{T}_{P}}$ is the time period of the bob to complete one rotation along the circle, the bob travels a distance of $2\pi R$ in time ${{T}_{P}}$. So the constant speed is, $v = \frac{{2\pi R}}{{{T_P}}}$. Also $R = L\sin \alpha $. Now putting the values of ${T_F} = \frac{w}{{\cos \alpha }}$ from Eq. \eqref{3} and $v = \frac{{2\pi R}}{{{T_P}}}$ in Eq. \eqref{4} and solving you get,

\[{T_P} = 2\pi \sqrt {\frac{R}{{g\tan \alpha }}} \tag{5} \label{5}\]

You know $R = L\sin \alpha $, so replacing $R$ in Eq. \eqref{5} by $L\sin \alpha $ you get,

\[{T_P} = 2\pi \sqrt {\frac{{L\cos \alpha }}{g}} \tag{6} \label{6}\]

Eq. \eqref{6} shows that time period of pendulum is related to the angle $\alpha$. When $\alpha$ increases the time period decreases.

Mechanics

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