Motion Vertical Circle

8 MCQs9-step worked example

Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

A bob on a string swings in a vertical circle. Energy conservation tells you the speed at each point — but energy alone does not tell you whether the string stays taut. That gap is the single high-frequency trap in this topic.

The setup. A bob of mass m hangs from a light, inextensible string of length L. At the lowest point it receives a horizontal velocity v₀. As it rises through angle θ from the vertical, it gains height h = L(1 − cos θ). By conservation of mechanical energy (NCERT Class 11 Physics Chapter 5, page 11):

½ m v₀² = ½ m v² + m g L(1 − cos θ)

so v² = v₀² − 2gL(1 − cos θ).

At the topmost point (θ = 180°), 1 − cos 180° = 2, giving v_top² = v₀² − 4gL.

Where students lose marks. Many aspirants stop here and write v₀² ≥ 4gL (speed at the top just reaches zero). This is incomplete. A string can only pull, not push — the tension T at the top must satisfy:

T + mg = m v_top² / L

For the string to remain taut, T ≥ 0, which requires v_top² ≥ gL. Combining with the energy equation: v₀² ≥ 5gL.

Key distinctions. For a string, the critical condition is v₀ = √(5gL). For a rigid rod, the bob only needs v_top ≥ 0, giving v₀ = √(4gL) — the rod can push. NEET questions specify "string" or "rod" deliberately; read that word.

At any angle θ, the tension is: T = (m v²/L) + mg cos θ = (m/L)(v₀² − 2gL + 3gL cos θ). Tension is maximum at the bottom (θ = 0°) and minimum at the top (θ = 180°).

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Direct ApplicationPractice

What is the minimum speed a bob of mass *m* must have at the lowest point of a vertical circle of radius *L* (string attached) to complete the full circle?

MCQ 2Direct ApplicationPractice

A bob on a string of length *L* moves in a vertical circle. If the speed at the lowest point is v₀ = √(5gL), what is the speed at the topmost point?

MCQ 3Easy RecallPractice

For a bob completing a vertical circle on a string of length *L*, the tension in the string is maximum at which position?

MCQ 4Direct ApplicationPractice

A bob on a rigid rod of length *L* is given a speed at the lowest point. What is the minimum speed at the bottom for the bob to reach the topmost point?

MCQ 5CalculationPractice

A 0.50 kg bob on a 1.0 m string moves in a vertical circle with speed √(5gL) at the bottom. Taking g = 10 m/s², what is the tension in the string at the lowest point?

MCQ 6Easy RecallPractice

A bob on a string of length *L* completes a vertical circle. At the topmost point, the tension in the string is zero. What is the speed of the bob at the topmost point?

MCQ 7Concept TrapPractice

A bob moves in a vertical circle on a string. At an angle θ from the lowest point (measured from the vertical), the speed is *v*. What is the expression for the tension in the string at that point?

MCQ 8CalculationPractice

A bob on a string is given speed v₀ at the bottom of a vertical circle of radius *L*. At what height above the lowest point does the bob have speed v₀/√2?

Worked Example

Pattern: Vertical circle — bob on string, minimum speed to complete the circle (NEET pattern: vertical circle bob string).

1
Given
- Mass m = 0.20 kg - String length (radius) L = 0.80 m - g = 10 m/s² (exact, problem-defined)
2
Required
Minimum speed v₀ at the lowest point such that the string remains taut throughout.
3
Concept
Two simultaneous constraints: (1) energy conservation connects bottom speed to top speed; (2) the string tension at the topmost point must be ≥ 0 (string can pull but not push).
4
Formula
Energy conservation (NCERT Class 11 Physics Chapter 5, page 11): ½mv₀² = ½mv_top² + mg(2L) Tension at the top: T_top + mg = mv_top²/L For minimum v₀, set T_top = 0: mg = mv_top²/L → v_top² = gL
5
Substitution
From v_top² = gL: v_top² = 10 × 0.80 = 8.0 m²/s² From energy conservation: ½v₀² = ½(8.0) + 10 × 2 × 0.80 ½v₀² = 4.0 + 16.0 = 20.0 v₀² = 40.0 m²/s²
6
Calculation
v₀ = √(40.0) = √(4 × 10) = 2√10 ≈ 6.3 m/s Note: g = 10 m/s² is an exact problem-defined value and the factor 2 in height (2L) is a geometric exact integer. Neither limits significant figures. The result 2√10 m/s is exact; the decimal approximation 6.3 m/s is rounded to 2 significant figures matching the given data (0.80 m).
7
Final answer
v₀_min = 2√10 m/s ≈ 6.3 m/s Equivalently: v₀² = 5gL = 5 × 10 × 0.80 = 40 m²/s².
8
Common trap
Using energy conservation alone gives v₀² ≥ 4gL = 32 m²/s² (v₀ ≈ 5.7 m/s). This is insufficient — the string goes slack before the bob reaches the top because the centripetal force requirement is not met. The extra gL term from the tension constraint is what aspirants miss.
9
Similar NEET-style question
A stone of mass 0.10 kg is tied to a string of length 0.50 m and whirled in a vertical circle. Find (a) the minimum speed at the lowest point to complete the circle, and (b) the tension in the string at the lowest point at this minimum speed. [Answer: (a) √(25) = 5.0 m/s; (b) 6mg = 5.9 N] ---

Before solving, remember these

Formula

Vertical circular motion — minimum speed at top

For a body of mass m tied to a string of length r and rotating in a vertical circle, the minimum speed at the topmost point (for the string to remain taut) is v_top = √(g r). From energy conservation, the corresponding speed at the bottom is v_bottom = √(5 g r).

-- NCERT Class 11 Physics, Ch. 5, p. 12

Formulas

10 formulas — click to collapse

Elastic collision — 1D final velocities

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}, v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}}

Final velocities of two bodies after an elastic head-on (1D) collision — momentum AND kinetic energy are both conserved. Special cases: equal masses exchange velocities; very heavy m2 at rest reflects m1 with reversed velocity.

Symbol	Quantity	SI Unit
m1, m2	Masses of the two bodies	kg
u1, u2	Initial velocities (signed, before collision)	m/s
v1, v2	Final velocities (signed, after collision)	m/s

Valid when

Collision is ELASTIC (kinetic energy conserved)
Head-on (1D) — for 2D collisions decompose along/perpendicular to line of impact

Do NOT use when

Inelastic collision (KE not conserved; use momentum-only + restitution)
Bodies stick together (perfectly inelastic case has its own formula)

Gravitational potential energy (near Earth)

U = m g h

Potential energy of a body of mass m at height h above the chosen reference level, in a region where g is approximately constant.

Symbol	Quantity	SI Unit
U	Gravitational PE	J
m	Mass	kg
g	Gravitational acceleration	m/s^2
h	Height above reference level	m

Valid when

Region small enough that g is uniform (typically near Earth's surface)
Reference level is freely chosen — only DIFFERENCES in U have physical meaning

Do NOT use when

Large altitude changes (use U = -GMm/r general form)
Below the reference level — h is signed

Kinetic energy

K = \frac{1}{2} m v^{2}

Energy a body possesses by virtue of its motion. Always non-negative. Frame-dependent through v.

Symbol	Quantity	SI Unit
K	Kinetic energy	J
m	Mass	kg
v	Speed	m/s

Valid when

Non-relativistic speeds (v << c)
Translational motion only (rotational KE = (1/2) I omega^2 separately)

Conservation of mechanical energy

K_{i} + U_{i} = K_{f} + U_{f}

If only conservative forces do work on a system, the total mechanical energy E = K + U is constant in time.

Symbol	Quantity	SI Unit
K_i, K_f	Initial, final kinetic energy	J
U_i, U_f	Initial, final potential energy	J

Valid when

Only CONSERVATIVE forces do work (gravity, spring, electrostatic — not friction or drag)
Closed system; no energy exchange with surroundings

Do NOT use when

Friction, drag, or other non-conservative forces do work
External forces add energy to the system

Perfectly inelastic 1D collision — common final velocity and KE loss

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}, Δ K = - \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

When two bodies stick together after a 1D collision, the common velocity is given by momentum conservation. The KE lost is converted to internal energy (heat, deformation).

Symbol	Quantity	SI Unit
v	Common final velocity	m/s
m1, m2	Masses	kg
u1, u2	Initial velocities	m/s
Delta_K	Change in kinetic energy	J

Valid when

Bodies stick together immediately after collision (perfectly inelastic)
Net external force = 0 during the brief collision (momentum conservation)

Instantaneous power

P = \frac{d W}{d t} = F \cdot v

Instantaneous power is the time rate of doing work; equivalently, the dot product of the force and velocity. Average power over an interval is W/t.

Symbol	Quantity	SI Unit
P	Power (instantaneous)	W
F	Force	N
v	Velocity	m/s

Valid when

F.v form when force and velocity are both known at the instant

Spring potential energy (Hooke's law regime)

U = \frac{1}{2} k x^{2}

Elastic potential energy stored in an ideal spring of force constant k that has been displaced by x from its natural length. Independent of the sign of x (compression or extension).

Symbol	Quantity	SI Unit
U	Elastic PE	J
k	Spring constant (force per unit displacement)	N/m
x	Displacement from natural length	m

Valid when

Spring obeys Hooke's law (F = -k*x) over the displacement range
No internal damping / hysteresis assumed

Do NOT use when

Beyond elastic limit (Hooke's law fails)
Real springs with finite damping (some elastic PE goes to internal energy)

Work done by a constant force

W = F \cdot s = F s cos θ

The work done by a constant force F on an object that undergoes a displacement s is the dot product F.s. Equivalently, W = (magnitude of F) * (magnitude of s) * cos(angle between them). Work is a scalar but has a sign.

Symbol	Quantity	SI Unit
W	Work (scalar, signed)	J
F	Constant force (vector)	N
s	Displacement (vector)	m
theta	Angle between F and s	rad/deg

Valid when

Force is CONSTANT in magnitude and direction over the displacement
Use the COMPONENT of force along the displacement, not magnitude alone

Do NOT use when

Force varies with position (use the variable-force integral)
Multiple forces act — apply this to each one separately or use net force

Work-energy theorem

W_{net} = Δ K = \frac{1}{2} m (v^{2} - v_{0}^{2})

The net work done by all forces on a particle equals the change in its kinetic energy. Holds for both constant and variable forces, in 1D and higher dimensions.

Symbol	Quantity	SI Unit
W_net	Net work done by all forces	J
Delta_K	Change in kinetic energy	J
m	Mass of particle	kg
v, v0	Final, initial speed	m/s

Valid when

W_net is the NET (vector-sum) work, not work of any one force
Particle (point-mass) idealisation; for extended bodies handle internal energy separately

Work done by a variable force

W = \int_{x_{i}}^{x_{f}} F (x) d x

When a force varies along the path, the work done is the line integral of force over displacement. In one dimension this is the area under the F-vs-x curve between the start and end positions.

Symbol	Quantity	SI Unit
W	Work done	J
F(x)	Force as a function of position	N
dx	Infinitesimal displacement	m
x_i, x_f	Initial and final positions	m

Valid when

Force may depend on position (e.g. spring force F = -k*x)
Generalises to W = integral F.dr in higher dimensions

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

12 items — click to collapse

Trap

Linear speed-distance ratio under uniform deceleration

Category: Overthinking

Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.

When it triggers

Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.

How to avoid

Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.

Trap

Incline-with-friction: missing μ cos θ term

Category: Sign Convention

Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.

When it triggers

Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.

How to avoid

On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.

Trap

Inelastic collision: defaulting to elastic

Category: Similar Terms

Student uses the elastic-collision velocity formulas (m1-m2)/(m1+m2) when the question explicitly says 'completely inelastic' (bodies stick).

When it triggers

Question says 'inelastic', 'stick together', 'after collision moves with common velocity'.

How to avoid

Perfectly inelastic 1D: v_common = (m₁ u₁ + m₂ u₂)/(m₁ + m₂). KE is NOT conserved; loss = ½ (m₁ m₂)/(m₁+m₂) × (u₁ - u₂)². Don't use elastic formulas.

Trap

Lift-power: friction term overlooked

Category: Overthinking

Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.

When it triggers

Question describes a lift moving at constant speed with explicit friction force on cable or guides.

How to avoid

At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.

Trap

Power problems: efficiency / loss factor dropped

Category: Overthinking

Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.

When it triggers

Question gives turbine, motor, or transformer with stated efficiency or loss percentage.

How to avoid

Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.

Trap

Instantaneous power: x vs v

Category: Similar Terms

Student plugs displacement x into P = F·v formula instead of velocity v.

When it triggers

Question gives displacement x(t) explicitly and a constant force; asks for instantaneous power.

How to avoid

P = F · v where v = dx/dt. Compute v first (differentiate x(t) once), THEN plug into F·v. P is NOT F·x.

Trap

Spring PE linear vs quadratic scaling

Category: Overthinking

Student treats spring PE as proportional to displacement (linear) instead of displacement-squared (quadratic). Common error: 'doubling the stretch doubles the PE'. Actual: doubling the stretch gives 4× the PE.

When it triggers

Question gives U at one stretch and asks for U at another. Distractors include linear-scaling answer (×2 instead of ×4 for double stretch).

How to avoid

U = ½ k x² is QUADRATIC. The PE-to-stretch ratio is the SQUARE of the stretch ratio: if stretch goes from x₁ to x₂, U_new / U_old = (x₂/x₁)².

Trap

Vertical circle: energy conservation alone is insufficient

Category: Overthinking

Student uses ½ m v₀² = m g (2L) (energy to reach top) and forgets the additional v_top² ≥ gL constraint for tension.

When it triggers

Question asks for minimum v₀ at lowest point so the bob can complete a full vertical circle.

How to avoid

TWO constraints: (1) energy: v_top² = v₀² - 4gL; (2) tension at top ≥ 0: v_top² ≥ gL. Combined: v₀² ≥ 5gL. Energy alone gives only v₀² ≥ 4gL which is insufficient.

Mistake

Student applies elastic-collision velocity formulas to an inelastic collision (or vice versa).

Root cause: concept gap

Correction

Elastic: BOTH momentum and kinetic energy conserved -> use the (m1-m2)/(m1+m2) form. Inelastic: ONLY momentum conserved; KE generally lost to heat. Perfectly inelastic: bodies stick together -> common velocity = (m1*u1 + m2*u2)/(m1+m2). Identify which type from the problem before choosing a formula.

Wrong option pattern

Distractor uses elastic-collision formulas for two bodies that the problem says stick together after impact.

Mistake

Student applies K_i + U_i = K_f + U_f in a problem where friction or other non-conservative forces clearly do work.

Root cause: formula misuse

Correction

Mechanical-energy conservation requires that ONLY conservative forces do work. When friction or drag is present, use the work-energy theorem directly: K_f - K_i = W_conservative + W_non-conservative, where W_non-conservative is typically negative (energy goes to heat).

Wrong option pattern

Distractor sets m*g*h = (1/2)*m*v^2 for a block sliding down a rough incline.

Mistake

Student treats gravitational potential energy as having an absolute value, instead of being defined relative to a chosen reference level.

Root cause: concept gap

Correction

PE is defined up to an additive constant. Only DIFFERENCES in PE have physical meaning (they equal the negative work done by the conservative force between two points). Choosing the ground as zero is a convention, not a derivation.

Wrong option pattern

Distractor offers a numerical PE value where two different reference levels would give different correct numbers.

Mistake

Student claims the normal reaction does positive (or negative) work on a block sliding on a horizontal surface.

Root cause: direction ignored

Correction

Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.