91影视

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,

$t_i=0s,{蝇}_i=-2rad/s$

And,

$t_f=4s,{蝇}_f=4rad/s$

Angular acceleration is the rate of change of angular velocity.

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

Using all the values,

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

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

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

We have

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

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

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

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