Statics and Dynamics Quiz: Conservation Of Energy

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Conservation Of EnergyQuestion 1 of 3

A pendulum consists of a bob of mass $m$ attached to a rigid, massless rod of length $L$ . The pendulum is released from a horizontal position (rod horizontal, bob at the same height as the pivot). Partway down, at the lowest point of the swing, the rod strikes a peg located a distance $d$ directly below the pivot, causing the effective pendulum length to suddenly change to $(L - d)$ .

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.

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)

d = \frac{3L}{5}

, which is derived from the string-pendulum condition requiring

v_{top}^2 = g(L-d)

, giving

mgL = mg\cdot 2(L-d) + \frac{1}{2}m \cdot g(L-d)

and solving for

d

d = \frac{2L}{5}

, derived by mistakenly placing the peg at distance

d

from the bob rather than from the pivot, so the effective new length is incorrectly taken as

d

instead of

(L-d)

d = L - \sqrt{\frac{2L}{5}}

, derived by conflating the rigid-rod and flexible-string conditions and applying a mixed energy equation that partially accounts for centripetal requirements at the top.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Conservation Of Energy

Practice Conservation Of Energy in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conservation Of Energy, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

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.

$d = L - \frac{L}{2} = \frac{L}{2}$ , 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 $mgL = mg \cdot 2(L-d)$ . (correct answer)
$d = \frac{3L}{5}$ , which is derived from the string-pendulum condition requiring $v_{top}^2 = g(L-d)$ , giving $mgL = mg\cdot 2(L-d) + \frac{1}{2}m \cdot g(L-d)$ and solving for $d$ .
$d = \frac{2L}{5}$ , derived by mistakenly placing the peg at distance $d$ from the bob rather than from the pivot, so the effective new length is incorrectly taken as $d$ instead of $(L-d)$ .
$d = L - \sqrt{\frac{2L}{5}}$ , derived by conflating the rigid-rod and flexible-string conditions and applying a mixed energy equation that partially accounts for centripetal requirements at the top.

Explanation: 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.

Question 2

A smooth bead of mass $m$ is threaded on a frictionless vertical circular loop of radius $R$ . The bead starts from rest at the bottom of the loop. A compressed spring at the bottom of the loop (oriented tangentially) gives the bead an initial kinetic energy $E_0$ by releasing its stored energy entirely into the bead before the bead leaves the spring.

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.

$E_0 = 2mgR$ , 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. (correct answer)
$E_0 = \frac{5}{2}mgR$ , because the bead must arrive at the top with a minimum speed satisfying $mg = mv_{top}^2/R$ , just as for a ball rolling on the inside of a loop-the-loop track.
$E_0 = \frac{5}{2}mgR - \frac{1}{2}mv_{top,min}^2$ , which simplifies to $2mgR$ only after substituting $v_{top,min}^2 = gR$ , yielding a contradictory result that reveals the minimum is actually $\frac{5}{2}mgR$ .
$E_0 = mgR$ , because the bead only needs to overcome half the gravitational potential energy change since the loop is circular and the average height gain is $R$ , not $2R$ .

Explanation: 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.

Question 3

A spacecraft of mass $M$ is in a circular orbit of radius $R$ around a planet of mass $m_p$ . The gravitational potential energy is $U = -Gm_pM/r$ , where $G$ is the gravitational constant and $r$ is the orbital radius. Mission control fires the engine briefly, reducing the spacecraft's speed by a small amount $\Delta v$ (so the new speed is $v_0 - \Delta v$ , where $v_0 = \sqrt{Gm_p/R}$ is the original circular-orbit speed). The spacecraft then follows an elliptical orbit.

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 $E = -Gm_pM/(2a)$ .

$a = \frac{R}{2 - R(v_0 - \Delta v)^2/(Gm_p)}$ , obtained by setting the total energy after the burn equal to $-Gm_pM/(2a)$ and solving for $a$ . (correct answer)
$a = \frac{R}{2 - Rv_0^2/(Gm_p)} = R$ , obtained by neglecting $\Delta v$ entirely and recovering the original circular orbit, which is the correct limiting case but not the answer for nonzero $\Delta v$ .
$a = \frac{R}{2 + R(v_0 - \Delta v)^2/(Gm_p)}$ , obtained by misapplying the vis-viva relation with a sign error, treating the kinetic energy term as positive in the denominator instead of negative.
$a = \frac{Gm_pM}{2(Gm_pM/R - \frac{1}{2}M(v_0-\Delta v)^2)}$ , obtained by writing the energy equation with $M$ in both numerator and denominator incorrectly, resulting in an expression that does not simplify to the vis-viva form.

Explanation: 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.