I 2nd derivative of angular displacement wrt time

If $\theta$ is angular displacement, does $\frac{d^2\theta}{dt^2} = (\frac{d\theta}{dt})^2$? Proof?

Parameterized, I think not. A contradiction:

$$\theta = sin(t)$$
$$\frac{d\theta}{dt} = cos(t)$$
$$\frac{d^2\theta}{dt^2} = -sin(t)$$
$$\left(\frac{d\theta}{dt}\right)^2 = (cos(t))^2$$

olivermsun

\begin{align} \theta(t) &= t \\ \frac{d\theta}{dt} &= 1 \\ \frac{d^2\theta}{dt^2} &= 0 \neq \left( \frac{d\theta}{dt} \right)^2 \end{align}

Yeah. Dimensionally they agree because $\theta$ is dimensionless, but they're not equivalent. Thanks.

FactChecker

Gold Member
2018 Award
Try a couple of sanity checks of the proposed equation:
1) If the rotation rate is not changing, the second derivative is zero. Does that mean that it is not rotating at all?
2) Since the right hand side is always positive, does that mean that the rotation rate can only get more positive?

• "2nd derivative of angular displacement wrt time"

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