Conservation of Energy - Statics and Dynamics

1. What is the minimum value of $E_0$ required for the bead to maintain contact with the loop at the very top? Note that for a bead on a wire (as opposed to a ball on a track), the wire can exert both inward and outward normal forces on the bead.
   Answer: $E_0 = 2mgR$, because the bead only needs enough energy to reach the top of the loop (height $2R$) with zero speed, since the wire can pull the bead inward if needed. When a bead is threaded on a wire loop, the wire can exert forces in both directions — inward and outward — unlike a ball on a track that can only be pushed inward by the surface. This distinction completely changes the minimum energy condition. For a ball on a frictionless track, contact is lost when the normal force drops to zero, requiring $v_{top}^2 \geq gR$ at the top. But for a bead on a wire, the wire can pull the bead inward (like a tension), so the bead stays on the loop regardless of speed — even at zero speed. The only requirement is that the bead physically reaches the top. Therefore, the minimum condition is simply $v_{top} = 0$, and energy conservation gives: $E_0 = \Delta PE = mg(2R)$ So the minimum initial kinetic energy is $E_0 = 2mgR$, confirming answer A. Answer B applies the condition $mg = mv_{top}^2/R$, which is the constraint for a ball on the inside of a track — not a bead on a wire. It incorrectly assumes the normal force can only push inward, leading to an unnecessarily large $E_0$. Answer C attempts an algebraic manipulation but starts from the wrong physical premise (the ball-on-track condition), so its "contradiction" is self-inflicted rather than revealing. Answer D misidentifies the height gained at the top as $R$ instead of $2R$, a geometric error — the top of a loop of radius $R$ is at height $2R$ above the bottom. Key tip: Always ask whether the constraint is one-sided (track) or two-sided (wire/rod). A bead on a wire or a ball on a rod changes the minimum speed condition dramatically — often dropping it to zero.
2. After the rod catches on the peg, what is the minimum distance $d$ such that the bob completes a full circular loop around the peg? Assume the rod is rigid (can push and pull), not a string.
   Answer: $d = L - \frac{L}{2} = \frac{L}{2}$, because for a rigid rod the bob only needs to reach the top of the new loop with zero velocity, requiring $(L-d) = L/2$ to satisfy the energy balance $mgL = mg \cdot 2(L-d)$. When a pendulum problem involves a rigid rod (not a string), the critical distinction is what condition must be met at the top of the loop. A string requires centripetal tension, so $v_{top}^2 \geq g(L-d)$. A rigid rod, however, can push as well as pull — meaning the bob only needs to arrive at the top with $v_{top} \geq 0$. This single insight unlocks the whole problem. Answer A is correct. At the moment the rod catches the peg (the lowest point), the bob has fallen a height $L$ from its release point, giving it kinetic energy $\frac{1}{2}mv^2 = mgL$. After catching the peg, the bob must rise a height $2(L-d)$ to reach the top of the new loop. Setting energy equal to zero velocity at the top: $mgL = mg \cdot 2(L-d)$, which gives $L = 2(L-d)$, so $L - d = \frac{L}{2}$ and $d = \frac{L}{2}$. Answer B applies the string condition $v_{top}^2 = g(L-d)$ to a rigid rod — this is the most common trap. Because the rod can push, you don't need any nonzero speed at the top, making B's derivation unnecessarily restrictive and yielding the wrong $d = \frac{3L}{5}$. Answer C misidentifies the geometry, treating $d$ as the new effective length rather than $(L-d)$. This is a setup misread that produces a completely incorrect energy equation. Answer D mixes the rigid-rod energy condition with leftover centripetal terms from the string case — a hybrid error that fits neither scenario. Study tip: Always identify whether the pendulum uses a rod or string first, since this determines your condition at the top. Rod → $v_{top} = 0$ is sufficient; String → $v_{top}^2 \geq g\ell$ is required.
3. Using conservation of energy and the vis-viva equation, what is the semi-major axis $a$ of the new elliptical orbit? Recall that for any orbit the total mechanical energy is $E = -Gm_pM/(2a)$.
   Answer: $a = \frac{R}{2 - R(v_0 - \Delta v)^2/(Gm_p)}$, obtained by setting the total energy after the burn equal to $-Gm_pM/(2a)$ and solving for $a$. When a spacecraft changes speed, its total mechanical energy changes, and that new energy uniquely determines the shape of the resulting orbit. The key tool here is recognizing that total mechanical energy $E = KE + PE$ can be equated to the orbital energy formula $E = -Gm_pM/(2a)$, letting you solve directly for the semi-major axis $a$. Right after the burn, the spacecraft is still at radius $R$, so its potential energy is unchanged at $-Gm_pM/R$. Its kinetic energy is now $\frac{1}{2}M(v_0 - \Delta v)^2$. Setting the total energy equal to $-Gm_pM/(2a)$: $\frac{1}{2}M(v_0-\Delta v)^2 - \frac{Gm_pM}{R} = -\frac{Gm_pM}{2a}$ Dividing through by $M$, rearranging, and solving for $a$ yields: $a = \frac{R}{2 - R(v_0-\Delta v)^2/(Gm_p)}$ This is exactly answer A — the correct result, cleanly derived by applying energy conservation with the actual post-burn speed. Answer B is a useful sanity check (when $\Delta v = 0$, you recover $a = R$), but it doesn't answer the question for nonzero $\Delta v$ — it simply ignores the burn altogether. Answer C contains a critical sign error: the kinetic energy term appears with a positive sign in the denominator instead of negative, which would imply a larger orbit when the spacecraft actually slows down and falls inward. Answer D retains $M$ in both numerator and denominator in a way that cancels incorrectly, producing an expression that looks dimensional but doesn't reduce to the standard vis-viva form. As a study tip: always cancel $M$ early when working orbital energy problems — if $M$ survives to the final answer for $a$, something has gone wrong.

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