91Ó°ÊÓ

In the given question , A hollow spherical shell has a mass of 8.20 kg and a radius of 0.220 m. It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890 rad/s 2.

We know that$K=\frac{1}{2}l{\mathrm{Ó\neg }}^2$ .

Here we need to find $\mathrm{Ó\neg }$and $l$, value.

We know that the angular acceleration,

role="math" localid="1664349159565" ${\mathrm{Ó\neg }}_z^2={\mathrm{Ó\neg }}_{z_0}^2+2{\mathrm{Î\pm }}_z\left(\mathrm{θ}-{\mathrm{θ}}_0\right)$___________(1)

In this question given,

$M=8.20,R=0.220.$

We know that this formula $l=\frac{2}{3}M{R}^2$.

$$\begin{gathered} =\frac{2}{3}\left(8.20kg\right){\left(0.220\right)}^2 \\ =0.2646kg.m^2 \end{gathered}$$

Put the value in equation (1)

$$\begin{gathered} {\mathrm{Ó\neg }}_z^2={\mathrm{Ó\neg }}_{z_0}^2+2{\mathrm{Î\pm }}_z\left(\mathrm{θ}-{\mathrm{θ}}_0\right) \\ {\mathrm{Ó\neg }}_z=\sqrt{2{\mathrm{Î\pm }}_z\left(\mathrm{θ}-{\mathrm{θ}}_0\right)} \end{gathered}$$

in the given question ${\mathrm{Î\pm }}_z\mathrm{θ}=0.890\frac{rad}{s^2}$.

$$\begin{gathered} =\sqrt{2\left(0.890\frac{rad}{s^2}\right)}\left(37.70red\right) \\ =8.192\frac{rad}{s} \end{gathered}$$

We know that$K=\frac{1}{2}l{\mathrm{Ó\neg }}^2$

$$\begin{gathered} K=\frac{1}{2}\left(0.2646kg.m^2\right)\left(8.192\frac{rad}{s}\right) \\ =8.88J \end{gathered}$$

Hence ,the kinetic energy of the shell after it has turned through 6.00rev is $8.88J$.