Name: Rotational Kinetic Energy
Uploaded: 2019-07-19
Duration: 10 min 15 s
Description: Do you think that rotational objects possess kinetic energy? Yes, as it is in motion. Then how do you calculate it? Using conventional formula! But there is...

Question

Deduce expression for kinetic energy of a body executing pure rotation in case of circular ring rotating about its diameter.

Accepted Answer

First, we will look at a ring about its axis passing through the centre

We will assume the mass of the ring to be M and radius be R.

Now we need to cut an elemental ring (dx) at the circumference of the ring. Hence, the mass (dm) of the elemental ring will be;

$d m = (m / 2 π R) d x$

Next, we calculate $d I = (d m) R^{2}$

$d I = [(m / 2 π R) d x] R^{2}$

Substituting the values, we get;

$d I = (m / 2 π R) R^{2} d x$

Using integration;

$\begin{array}{rcl} I & = & (m R / 2 π) \int_{0}^{2 π R} d x \\ = & (m R / 2 π) {[x]}_{0}^{2 π R} \\ = & m R^{2} \end{array}$

For the second expression, we will be dealing with the moment of inertia of a ring about an axis passing through its diameter.

Here we will recall the formula;

$I = \int r'^{2} d m$

We need to find $d m$ . We will get;

$d m = (M / 2 π) d θ$

Meanwhile, $r' = r \cos θ$

The next step is to carry out the integration. We will have;

$\begin{array}{rcl} I & = & \int_{0}^{2 π} r^{2} \cos^{2} θ (M / 2 π) d θ \\ = & (M r^{2} / 2 π) \int_{0}^{2 π} \cos^{2} θ d θ \\ = & (M r^{2} / 2 π) {[\frac{θ}{2} + \frac{\sin^{2} θ}{4}]}_{0}^{2 π} \\ = & \frac{1}{2} M r^{2} \end{array}$

The formula to calculate the kinetic energy of the ring is given by

$K = \frac{1}{2} I ω^{2} . . . (1)$

Substitute the expression for the moment of inertia into equation (1) and simplify to obtain the required kinetic energy.

$\begin{array}{rcl} K & = & \frac{1}{2} (\frac{1}{2} M r^{2}) ω^{2} \\ = & \frac{1}{4} M r^{2} ω^{2} \end{array}$

Deduce expression for kinetic energy of a body executing pure rotation in case of circular ring rotating about its diameter.

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