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Physics of Ground Spinner (Chakkar) - Learn Physics on Diwali


The ground spinner (or Chakkar, Chakhri, Firki) is one of the firecracker that made our childhood during Diwali. We cannot forget the act of jumping on the sparks as soon as it starts. The ground spinner consists of a propellant (gunpowder) filled tube wound in the form of a spiral. When ignited, the Chakkar starts rotating with an ever increasing angular speed. Let us learn more about it through a simplified model and an interesting problem.

Simplified Model and the Problem

Let propellant filled tube is of the length $l_0=30$ cm and width $D=0.5$ cm. The tube burns at a constant rate of $\lambda=1$ cm/s. The tangential thrust at the end of the tube is $F=1$ N. The total mass of the tube is $m_0=30$ gram. What is the angular speed of the spinner at time $t$? Assume that spinner rotate about a frictionless fixed axis passing through its centre. The radius of the spiral of length $l$ can be approximated as $r=\sqrt{4\pi D l}$. You are free to make appropriate approximations like uniform mass distribution etc.

Ground Spinner - Chakkar

Solution and Discussion

Let us make a qualitative analysis. As time progresses, the length, radius and mass of the spiral starts decreasing. Thus, moment of inertia of the spiral about the axis of rotation decreases with time. Also, torque about the axis of rotation decreases with time due to decrease in radius.

The length $l$, radius $r$ and mass $m$ at time $t$ (after ignition) are given by \begin{align} l&=l_0-\lambda t\\ r&=\sqrt{4\pi Dl}\\ &=\sqrt{4\pi D (l_0-\lambda t)} \\ m&=\frac{m_0}{l_0}(l_0-\lambda t) \end{align} The torque about the axis of rotation due to the tangential force $F=1$ N acting at the end of the tube is \begin{align} \tau&=rF\\ &=\sqrt{4\pi D(l_0-\lambda t)} F \end{align} We approximate spiral as a uniform solid disc of mass $m$ and radius $r$. The moment of inertia of the spinner at time $t$ is given by \begin{align} I&=\frac{1}{2}mr^2\\ &=\frac{1}{2}\;\frac{m_0}{l_0}(l_0-\lambda t) \; 4\pi D(l_0-\lambda t) \\ &=\frac{2\pi m_0 D}{l_0} (l_0-\lambda t)^2 \end{align} The angular acceleration of the spinner at time $t$ is given by \begin{align} \alpha &=\frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{\tau}{I}\\ &=\frac{l_0 F}{m_0}\sqrt{\frac{1}{\pi D}}\;(l_0-\lambda t)^{-3/2} \end{align} Integrate with initial condition $\omega=0$ at $t=0$ to get \begin{align} \omega&=\frac{l_0 F}{m_0}\sqrt{\frac{1}{\pi D}}\int_0^t (l_0-\lambda t)^{-3/2} \mathrm{d}t \\ &=\frac{2F}{m_0 \lambda} \sqrt{\frac{l_0}{\pi D}}\left[ \frac{1}{\sqrt{1-\lambda t/l_0}} -1 \right] \end{align} The total time to burn is $T=l_0/\lambda$. When time is small in comparison to the total burn time (i.e., $t \ll l_0/\lambda$) then above expression can be approximated as (using binomial expansion) \begin{align} \omega=\frac{F}{m_0}\sqrt{\frac{1}{\pi D l_0}} t \end{align} The angular speed increase linearly with time which seems reasonable.

However, the angular speed becomes infinite if we substitute $t=l_0/\lambda$. This cannot be correct. Often, oversimplified models (as we used in this problem) leads to unrealistic results.

I encourage you to substitute values of parameters to get an estimate of angular velocity.

Let angular velocity at the end of burning is $\omega_m$. Can you estimate the time taken by the spinner to come to rest? The enjoyment starts when the spinner is ignited and it finishes when it comes to rest.

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