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. What is the kinetic energy of the shell after it has turned through 6.00 rev?

Short answer

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

Step-by-step solution

Step:-1 explanation

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.

Step:-2 concept  

We know that$K=\frac{1}{2}l{\omega }^2$ .

Here we need to find $\omega$and $l$, value.

We know that the angular acceleration,

role="math" localid="1664349159565" ${\omega }_z^2={\omega }_{z_0}^2+2{\alpha }_z\left(\theta -{\theta }_0\right)$___________(1)

Step:-3 calculation

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)

$$\begin{gathered} {\omega }_z^2={\omega }_{z_0}^2+2{\alpha }_z\left(\theta -{\theta }_0\right) \\ {\omega }_z=\sqrt{2{\alpha }_z\left(\theta -{\theta }_0\right)} \end{gathered}$$

in the given question ${\alpha }_z\theta =0.890\frac{rad}{s^2}$.

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

We know that$K=\frac{1}{2}l{\omega }^2$

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

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