In the given figure $L = 1.25 \mathrm m$ and $\theta = 40^\circ$ What is the least value that $v_\circ$ can have for the bob of the pendulum to swing down and then swing up to a) a horizontal position, b) a vertical position with cord remaining straight ?
(Sorry for this monster pic I don't intend to do this but I have no clue as to how to use Inkscape.)
My understanding of the question tells me that i have to find the minimum $v_0$ for the pendulum to reach the red positions.
For part a) I did the following,
Reference point for the potential energy : - Lowest position the pendulum.
System :- Earth and pendulum.
Initial height of the pendulum from the reference point = $L - L\cos 40 \approx 0.3$
Answer:-
$$\begin{align}\\ 0 &= \triangle K + \triangle U \\&=\frac 12 m(0 - v_0^2) + mg(1.25 - 0.3)\end{align}$$
$$\implies\frac 12 v_0^2 = g \times 0.95 \implies v_0 = 4.31 \mathrm {m/s}$$
For part b) I did the following :-
Reference point for the potential energy : - Lowest position the pendulum.
System :- Earth and pendulum.
Answer:-
$$\begin{aligned} 0 &= \triangle K + \triangle U \\ &=\frac 12 m(0 - v_0^2) + mg(2.5 - 0.3)\end{aligned}$$
$$\implies\frac 12 v_0^2 = g \times 2.2 \implies v_0 = 6.5 \mathrm {m/s}$$
My answer to (b) is incorrect :( .
Calculation given in the book :-
Let $u = \sqrt{Lg}$
$$\begin{aligned} 0 &= \triangle K + \triangle U \\ &=\color{red}{\frac 12 mu^2} - \frac 12 mv_0^2 + mg(2.5 - 0.3)\end{aligned}$$
$$\implies\frac 12 v_0^2 = g \times 2.2 + {g\times 1.25 \over 2} \implies v_0 = 7.45 \mathrm {m/s}$$
My confusions:
I know $u$ is the final velocity of the pendulum and the highest point but does not know why we considered it and how did we get its value?
Should not the velocity of the bob at the highest point is 0 and the KE 0 for minimum value of $v_0$?
Why did not we get a similar term for question a)?