91Ó°ÊÓ

Given that the angular acceleration is not constant.

The expression for the rotational kinetic energy is given by,

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}{\mathit{Ó\neg }}^{\mathbf{2}}$

Here, lis the moment of inertia, and$\mathit{Ó\neg }$is the angular velocity of body.

The equation$v=r\mathrm{Ó\neg }$is obtained from differentiation of the equation$s=r\mathrm{θ}$.

Here, v is the linear velocity, r is the distance, s is the arc distance, and$\mathrm{θ}$is the angle.

The equation$s=r\mathrm{θ}$ is valid for both inertial and accelerating frames of reference.

Here, it does not matter whether angular acceleration is constant or not. Hence, the equation $v=r\mathrm{Ó\neg }$is valid.

The tangential acceleration is$a_{\tan }=r\mathrm{Î\pm }$.

Here,$\mathrm{Î\pm }$is the angular acceleration.

When an object moves along a curve, it possesses tangential acceleration irrespective of the rate of increase of the speed along the curve. Hence, the equation localid="1667973436097" $a_{\tan }=r\mathrm{Î\pm }$is valid.

he final angular velocity is given by,

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$

Here, ${\mathrm{Ó\neg }}_0$is the Initial angular velocity and t is the time.

The equation $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$is obtained by assuming angular acceleration of the object is constant.

Hence, the equation $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$is not valid.

The tangential acceleration is given by,

$a_{\tan }=r{\mathrm{Ó\neg }}^2$

When an object moves along a circular path, it possesses centripetal acceleration irrespective of the rate of increase of the speed along the curve.

Hence, the equation $a_{\tan }=r{\mathrm{Ó\neg }}^2$is valid.

The kinetic energy is given by,

$K=\frac{1}{2}m{v}^2$

Here, m is mass.

The equation$K=\frac{1}{2}m{v}^2$is obtained by using the equations$K=\frac{1}{2}m{v}^2$and$v=r\mathrm{Ó\neg }$.

$$\begin{gathered} K=\frac{1}{2}m{(r\mathrm{Ó\neg })}^2=\frac{1}{2}m{r}^2{\mathrm{Ó\neg }}^2 \\ =\frac{1}{2}l{\mathrm{Ó\neg }}^2 \end{gathered}$$

Since$v=r\mathrm{Ó\neg }$is valid for any accelerationrole="math" localid="1663935915591" $K=\frac{1}{2}l{\mathrm{Ó\neg }}^2$is valid as well.

Hence, the equation role="math" localid="1663935944634" $K=\frac{1}{2}{l\mathrm{Ó\neg }}^2$is valid.