91Ó°ÊÓ

The derivative of the angular position function with respect to time is the instantaneous angular velocity.

The change in an object's angular velocity with respect to time is known as average angular acceleration.

Amplitude $\mathrm{θ}\left(t\right)$ is expressed as,

$\mathrm{θ}\left(t\right)=\mathrm{Θ}cos(\mathrm{Ó\neg }t+\mathrm{Ï•})$

Here, $\mathrm{Θ}$ is the angular amplitude, $\mathrm{Ó\neg }$ is the angular frequency, t is time, and $\mathrm{Ï•}$ is the phase.

The phase $\mathrm{Ï•}$ is zero here. Thus evaluating the expression for angular velocity,

$\begin{array}{r}\frac{d\mathrm{θ}}{dt}=\mathrm{Θ}\frac{d}{dt}\cos \left(\mathrm{Ó\neg }t\right)\\ =−\mathrm{Ó\neg }\mathrm{Θ²õ¾\pm ²Ô}\left(\mathrm{Ó\neg }t\right)\end{array}$

And the expression of the angular acceleration,

$\begin{array}{r}\mathrm{Î\pm }=\frac{{\mathrm{d}}^2\mathrm{θ}}{{\mathrm{dt}}^2}\\ =\mathrm{Θ}\frac{\mathrm{d}}{\mathrm{dt}}\left(\cos \mathrm{Ó\neg }\mathrm{t}\right)\\ =\frac{\mathrm{d}}{\mathrm{dt}}(−\mathrm{Ó\neg }\mathrm{Θ²õ¾\pm ²Ô}(\mathrm{Ó\neg }\mathrm{t}\left)\right)\\ =−\mathrm{Ó\neg }\mathrm{Θ}(−\mathrm{Ó\neg }\cos (\mathrm{Ó\neg }\mathrm{t}\left)\right)\\ \mathrm{Î\pm }=−{\mathrm{Ó\neg }}^2\mathrm{Θ³¦´Ç²õ}\left(\mathrm{Ó\neg }\mathrm{t}\right)\end{array}$

Hence, the expression for angular velocity $d\mathrm{θ}/dt$ and angular acceleration $d^2\mathrm{θ}/{\mathrm{dt}}^2$ as functions of time are $-\mathrm{Ó\neg }\mathrm{Θ}\sin \left(\mathrm{Ó\neg }t\right)$ and $-{\mathrm{Ó\neg }}^2\mathrm{Θ}\cos \left(\mathrm{Ó\neg }t\right)$ respectively.

When the displacement is given to be $\mathrm{Θ}$, then,

$\mathrm{Θ}=\mathrm{Θ}=\mathrm{Θ³¦´Ç²õ}\left(\mathrm{Ó\neg }\mathrm{t}\right)$= which occurs at t = 0 .

So, role="math" localid="1668149421316" $\mathrm{Î\pm }=-{\mathrm{Ó\neg }}^2\mathrm{Θ}$

When the displacement is given to be $\mathrm{Θ}/2$, then,

$\begin{array}{l}\frac{\mathrm{Θ}}{2}=\mathrm{Θ}cos\left(\mathrm{Ó\neg }t\right)\\ \frac{1}{2}=cos\left(\mathrm{Ó\neg }t\right)\end{array}$

So, the angular velocity is given as,

$\frac{d\mathrm{θ}}{dt}=\frac{-\mathrm{Ó\neg }\mathrm{Θ}\sqrt{3}}{2}$

And acceleration is given as

$\mathrm{Î\pm }=\frac{-{\mathrm{Ó\neg }}^2\mathrm{Θ}}{2}$

Hence, the balance wheel’s angular velocity and angular acceleration are$\frac{-\mathrm{Ó\neg }\mathrm{Θ}\sqrt{3}}{2}$and$\mathrm{Î\pm }=\frac{-{\mathrm{Ó\neg }}^2\mathrm{Θ}}{2}$respectively.