Figure 10-33gives angular speed versus time for a thin rod that rotates around one end. The scale on the v axis is set by $\mathit{\omega }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$(a) What is the magnitude of the rod’s angular acceleration? (b) At t = 4.0s , the rod has a rotational kinetic energy of 1.60J. What is its kinetic energy at t = 0?

By using the concept of Angular acceleration, Rotational kinetic energy and moment of inertia we can find angular acceleration and rotational kinetic energy at t = 0 s. The formulas used for the concept are given below.

From the graph,

At,

$t_i=0s,{\omega }_i=-2rad/s$

And,

$t_f=4s,{\omega }_f=4rad/s$

Angular acceleration is the rate of change of angular velocity.

$\begin{array}{rcl}\alpha & =& \frac{d\omega }{dt}\\ & =& \frac{{\omega }_f-{\omega }_i}{t_f=t_i}\end{array}$

Using all the values,

$\begin{array}{rcl}\alpha & =& \frac{4rad/s-\left(-2rad/s\right)}{4s-0}\\ & =& \frac{6}{4}rad/s^2\\ & =& 1.5rad/s^2\end{array}$

Hence the value of angular acceleration is, 1.5rad/s².

Now to solve part b, we will calculate the inertia of the thin rod using rotational kinetic energy $K=1.60Jatt=4s$

Rearranging rotational kinetic energy formula for inertia I

$\begin{array}{rcl}K& =& \frac{1}{2}l{\omega }^2\\ l& =& \frac{2K}{{\omega }^2}\end{array}$

We have

$\begin{array}{rcl}att& =& 4s,K=1.60J,\omega =4rad/s\\ l& =& \frac{2\times 1.60J}{{\left(4rad/s\right)}^2}\\ & =& 0.2kg.m^2\end{array}$

Rotational inertia of the thin rod (I) is 0.2kh.m² , we will use this to find rotational kinetic energy at,

$\begin{array}{rcl}t& =& 0s,\omega =-2rad/s\\ K& =& \frac{1}{2}l{\omega }^2\\ & =& \frac{1}{2}\times 0.2kg.m^2\times {\left(-2rad/s\right)}^2\\ & =& 0.4J\end{array}$

Hence the kinetic energy of the thin rod at t = 0 is 0.4J.