Answer
a) $\omega=3.52rad/s$
b) $E=2.04\times10^{-2}J$
c) $v=0.41m/s$
Work Step by Step
a) The angular frequency of the motion is $$\omega=\sqrt{\frac{g}{L}}=\sqrt{\frac{9.8m/s^2}{0.79m}}=3.52rad/s$$
b) Because total mechanical energy is conserved throughout the motion, we can calculate the total mechanical energy at the initial point, before the pendulum is released.
Initially, the pendulum does not have any speed, so $KE_0=0$. Therefore,
$$E_0=KE_0+PE_0=0+mgh_0$$
The particle's weight $mg=0.24\times9.8=2.35N$.
From the figure below, we see that $h_0-h_f=L-L\cos8.5=0.79-0.79\cos8.5=8.68\times10^{-3}m$. If we take $h_f=0$, then $h_0=8.68\times10^{-3}m$
$$E_0=2.04\times10^{-2}J$$
c) The lowest point is the reference point, so $PE=0$. Because energy is conserved, $$KE+PE=\frac{1}{2}mv^2+0=2.04\times10^{-2}J$$ $$v=\sqrt{\frac{2.04\times10^{-2}J}{\frac{1}{2}m}}=\sqrt{\frac{2.04\times10^{-2}J}{0.12kg}}$$ $$v=0.41m/s$$
