Solution

Let position A be the reference point for potential energy, $$U_{A} = 0$$. The total mechanical energies at A, B, and C are:

$$E_{A} = \dfrac{1}{2} mv^2_{A} + U_{A} = \dfrac{1}{2} mv^2_0$$

$$E_{B} = \dfrac{1}{2} mv^2_{B} + U_{B} = \dfrac{1}{2} mv^2_{B} - mgL$$

$$E_{D} = \dfrac{1}{2} mv_{D}^2 + U_{D} = mgL$$

where $$v_{D} = 0.$$ The problem can be analyzed by applying energy conservation:

$$E_{A} = E_{B} = E_{D}.$$

(a) The condition $$E_{A} = E_{D}$$ gives

$$\dfrac{1}{2} mv^2_0 = mgL \Rightarrow v_0 = \sqrt{2gL}.$$

(b) To find the tension in the rod when the ball passes through B, we first calculate the speed at B. Using $$E_{B} = E_{D},$$ we find

$$\dfrac{1}{2} mv^2_{B} - mgL = mgL \Rightarrow v_{B} = \sqrt{4gL}.$$

The direction of the centripetal acceleration is upward (at that moment), as is the tension force. Thus, Newton’s second law gives

$$T - mg = \dfrac{mv^2_B}{r} = \dfrac{m(4gL)}{L} = 4mg$$
or $$T = 5mg.$$

(c) The difference in height between C and D is L, so the “loss” of mechanical energy(which goes into thermal energy) is $$-mgL.$$

(d) The difference in height between B and D is 2L, so the total “loss” of mechanical energy (which all goes into thermal energy) is $$-2 mgL.$$

Note: An alternative way to calculate the energy loss in (d) is to note that
$$E'_{B} = \dfrac{1}{2} mv'^2_{B} + U_{B} = 0 - mgL = -mgL$$
which gives
$$\Delta E = E'_{B} - E_{A} = -mgL - mgL = -2mgL.$$