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When we write the expression for the kinetic Energy of a rotating body, we write is as $$E = \frac{I_{\text{CM}}{\omega}^2}{2} + \frac{M{V_{\text{CM}}}^2}{2}$$

So, basically we wrote the expression as $$E_{\text{body/ground}} = E_{\text{body/CM}} + E_{\text{CM/ground}}$$

However, I want to consider a case where I write the expression about a point which is at rest. For example, if I take a cylinder rolling without slipping on a surface, can I write it as $$E= \frac{I_{\text{IAOR}}{\omega}^2}{2}$$ where IAOR is the instantaneous axis of rotation, because it is at rest?

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    $\begingroup$ Yes, since motion wrt IAOR will be purely rotational, so translation KE will be zero $\endgroup$
    – paul230_x
    Jul 8, 2021 at 4:22

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This is simple to see how the equation holds mathematically, $$E= K.E._{trans} + K.E._{rot} \\ \implies E= \frac{I_{\text{CM}}{\omega}^2}{2} + \frac{M{V_{\text{CM}}}^2}{2} \\ \implies E = \frac{(I_{\text{CM}} + MR_{CM}^2){\omega}^2}{2} \\ \implies E=\frac{I_{IOAR}\omega^2}{2}$$

since about the IOAR $V_{CM}= R_{CM}\omega $and $I_{IOAR} = I_{CM} + MR_{CM}^2$ where $R_{CM}$ is the distance of center of mass from the instantaneous axis of rotation.

You could also say that it is purely rotational motion so you can use $E=\frac{I\omega^2}{2}$ but you cannot say it is at rest.

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  • $\begingroup$ I meant that the IAOR is at rest for the instant. $\endgroup$ Jul 8, 2021 at 12:39

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