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. 79Motion Vertical Circle
8 MCQs2 revision cards9-step worked example
Source: NCERT Work, Energy and PowerOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Motion Vertical Circle, explained for NEET
The trap here is finishing the problem one constraint too early. Asked for the minimum launch speed at the bottom of a vertical circle, most repeaters write energy conservation, get v₀² = 4gL, and stop. That answer is wrong, and it is wrong for a reason worth holding onto: arriving at the top is not the same as going round.
Energy conservation tells you the speed at any height. With the lowest point as reference and the top at height 2L, mechanical energy conservation (NCERT Class 11 Physics Chapter 5, page 82) gives
v_top² = v₀² − 4gL
Set v_top = 0 and you get v₀² = 4gL — enough to reach the top with nothing left over. But a string cannot push. At the top, both gravity and tension point downward toward the centre, so the circular-motion requirement is
T + mg = m v_top² / L
Tension can only be pulled taut, never compressed: T ≥ 0. Putting T = 0 in that equation gives the floor on speed at the top:
v_top² ≥ gL
Combine the two. v₀² − 4gL ≥ gL, so v₀² ≥ 5gL, and the minimum launch speed is v₀ = √(5gL).
Between 4gL and 5gL the bob does something the energy equation never mentions: the string goes slack partway up, the bob leaves the circular path, and it flies as a projectile. The circle is only maintained while the string stays taut.
Two consequences for NEET. First, a rigid rod changes the answer — a rod can push, so T ≥ 0 does not apply and v_top may be zero; the rod condition is v₀² ≥ 4gL. Read the wording. Second, at the lowest point tension is largest (T − mg = m v₀²/L), giving T = 6mg at the minimum-speed condition — a favourite follow-up.
Watch-out: whenever a vertical-circle question says "just completes," ask yourself which constraint is binding — energy, or the string.
Can you answer these Motion Vertical Circle MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
For a bob on a light string moving in a vertical circle of radius L, the minimum speed at the highest point needed for the string to remain taut is:
Show answer and why every option is right or wrong
Answer: B. At the top, tension and weight both point toward the centre, so T + mg = mv²/L. The taut limit is T = 0, giving v_top = √(gL). Only the centripetal condition at the top is needed here; the mechanical-energy relation on page 82 of NCERT Class 11 Physics Chapter 5 is what carries this to the lowest point, where it gives √(5gL).
Why A is wrong: A is wrong because √(2gL) is the speed gained by falling freely through a height L; it is not the tension-limited condition at the top.
Why C is wrong: C is wrong because √(5gL) is the minimum speed at the LOWEST point, not the top. The question asks for the top.
Why D is wrong: D is wrong because a string cannot push. Zero speed at the top would require an upward force to hold the bob on the circle, which a string cannot supply — this is exactly the energy-only conclusion the tension constraint corrects.
MCQ 2Easy RecallPractice
In the equation T + mg = mv_top²/L for a bob at the top of a vertical circle, the two terms on the left are added rather than subtracted because:
Show answer and why every option is right or wrong
Answer: D. The centripetal equation sums force components pointing toward the centre. At the highest point the centre lies directly below the bob, so the downward weight and the downward string tension both count positively.
Why A is wrong: A is wrong because at the minimum-speed condition tension at the top is exactly zero, which is smaller than mg, not larger.
Why B is wrong: B is wrong because kinetic energy being minimum at the top is true but irrelevant to the direction of the force terms; the sign comes from geometry, not from energy.
Why C is wrong: C is wrong because the bob is not at rest at the top — if it were, the string would already have gone slack on the way up.
MCQ 3Easy RecallPractice
A bob on a light string is to complete a full vertical circle of radius L. The minimum speed required at the lowest point is:
Show answer and why every option is right or wrong
Answer: B. Energy conservation gives v_top² = v₀² − 4gL, and the taut-string condition at the top requires v_top² ≥ gL. Combining the two yields v₀² ≥ 5gL, so v₀ = √(5gL).
Why A is wrong: A is wrong because √(2gL) would not even carry the bob to a height of 2L; it corresponds to a rise of only L.
Why C is wrong: C is wrong because √(6gL) exceeds the requirement; the question asks for the minimum, and 6gL is larger than the binding value 5gL.
Why D is wrong: D is wrong because √(4gL) is the energy-only answer: enough to arrive at the top with zero speed, but the string goes slack before that and the bob leaves the circle.
MCQ 4Direct ApplicationPractice
A bob of mass m on a light string of length L is moving in a vertical circle. At the instant it passes the lowest point with speed v, the tension in the string is:
Show answer and why every option is right or wrong
Answer: A. At the lowest point the centre of the circle is directly above the bob, so tension acts toward the centre (upward) and weight away from it (downward): T − mg = mv²/L, giving T = mv²/L + mg.
Why B is wrong: B is wrong because it subtracts the weight instead of adding it — this is the result of copying the top-of-circle sign convention to the bottom without redrawing the directions.
Why C is wrong: C is wrong because it reverses the whole equation; tension at the lowest point exceeds mg, so this expression would be negative for any v² > gL.
Why D is wrong: D is wrong because it omits the weight entirely, treating the string as though it only had to supply the centripetal force with no gravity to oppose.
MCQ 5Direct ApplicationPractice
A bob on a light string of length L is launched horizontally from the lowest point at exactly the minimum speed for a complete vertical circle. The tension in the string at the lowest point at that instant is:
Show answer and why every option is right or wrong
Answer: C. At the minimum condition v₀² = 5gL. Substituting into T = mv₀²/L + mg gives T = 5mg + mg = 6mg.
Why A is wrong: A is wrong because zero tension occurs at the TOP under the minimum condition, not at the bottom where tension is greatest.
Why B is wrong: B is wrong because mg is the weight term alone; it ignores the centripetal contribution mv₀²/L = 5mg.
Why D is wrong: D is wrong because 5mg is the centripetal term mv₀²/L on its own, with the weight term mg dropped from the sum.
MCQ 6Direct ApplicationPractice
A small bob on a light string of length 0.40 m swings in a vertical circle. Taking g = 10 m/s² (exact), the minimum speed at the lowest point for the string to stay taut throughout is:
Show answer and why every option is right or wrong
Answer: C. v₀ = √(5gL) = √(5 × 10 × 0.40) = √20 = 4.47 m/s, which to two significant figures is 4.5 m/s.
Why A is wrong: A is wrong because 2.8 m/s is √(2gL) — the free-fall speed after dropping a height L, not the vertical-circle condition.
Why B is wrong: B is wrong because 4.0 m/s is √(4gL), the energy-only answer that ignores the requirement that the string carry non-negative tension at the top.
Why D is wrong: D is wrong because 2.0 m/s is √(gL), the speed required at the TOP; the question asks for the speed at the lowest point.
MCQ 7Concept TrapPractice
A bob on a light string of length L is given a speed at the lowest point such that 4gL < v₀² < 5gL. Which description of the subsequent motion is correct?
Show answer and why every option is right or wrong
Answer: D. Energy alone permits the bob to reach the top in this range, but the taut condition v² ≥ gL at the top fails. Tension reaches zero somewhere above the horizontal, and once the string is slack the only force is gravity, so the bob follows a projectile path.
Why A is wrong: A is wrong because the circle is not completed at all in this range. Completing it requires v₀² ≥ 5gL, so no speed below that maintains circular motion to the top.
Why B is wrong: B is wrong because a bob at rest at the top with a slack or taut string is not in equilibrium — gravity is unbalanced, so it cannot remain there.
Why C is wrong: C is wrong because v₀² > 4gL is more than enough energy to rise past the horizontal; oscillation below the horizontal occurs only for v₀² ≤ 2gL.
MCQ 8CalculationPractice
The bob on a light string is replaced by an identical bob on a light rigid rod of the same length L, pivoted at the same point. The minimum speed at the lowest point for a complete vertical circle changes from √(5gL) to:
Show answer and why every option is right or wrong
Answer: A. A rigid rod can push as well as pull, so the constraint T ≥ 0 no longer applies and the rod may support the bob from below at the top. The only surviving requirement is that the bob reach the top, i.e. v_top ≥ 0, which from v_top² = v₀² − 4gL gives v₀ = √(4gL).
Why B is wrong: B is wrong because √(5gL) is derived from the string's inability to push. Removing that inability removes the gL margin at the top.
Why C is wrong: C is wrong because √(gL) is the speed needed AT the top for a string, not the launch speed at the bottom for a rod.
Why D is wrong: D is wrong because √(3gL) corresponds to a rise of 1.5L, which is short of the top at height 2L; the bob would not get there at all.
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Motion Vertical Circle: quick recall before you leave
How do you solve a Motion Vertical Circle question? A worked example
- 1
Given.
Mass of bob
m = 0.20 kg.
String lengthL = 1.0 m(light, inextensible).g = 10 m/s²(exact, as stated in the problem).
The bob is given a horizontal speedv₀at the lowest point. - 2
Required.
(a) The minimum
v₀for the bob to complete a full vertical circle with the string taut throughout.
(b) The tension in the string at the lowest point at that minimum speed. - 3
Concept.
Two independent conditions must both be satisfied. Mechanical energy is conserved because tension acts perpendicular to the velocity everywhere and does no work, while gravity is conservative — this fixes the speed at the top for a given
v₀. Separately, the string can only pull, so the tension at the top must be non-negative — this sets a floor on the speed at the top. The binding condition is whichever is stricter, and it is the tension one. - 4
Formula.
Energy, lowest point to top (height
2L):½ m v₀² = ½ m v_top² + m g (2L)→v_top² = v₀² − 4gL
Centripetal equation at the top, both forces toward the centre:T_top + mg = m v_top² / L
Centripetal equation at the bottom, tension toward the centre and weight away:T_bottom − mg = m v₀² / L - 5
Substitution.
Minimum condition: set
T_top = 0.mg = m v_top²/L→v_top² = gL
Feed into the energy relation:gL = v₀² − 4gL→v₀² = 5gL
Numerically:v₀² = 5 × 10 × 1.0 = 50 m²/s²
For the tension at the bottom:T_bottom = m v₀²/L + mg = 0.20 × 50/1.0 + 0.20 × 10 - 6
Calculation.
v₀ = √50 = 7.07… m/sT_bottom = 10 + 2.0 = 12 N
Note on exact quantities:g = 10 m/s²is given as exact by the problem, and the factors 5, 4 and 2 in these relations are counting/geometric constants from the derivation. None of them limits the significant-figure count. The precision is set by the measured inputsm = 0.20 kgandL = 1.0 m, both two significant figures, so the answers carry two. - 7
Final answer.
(a)
v₀ = 7.1 m/s(2 s.f.)
(b)T_bottom = 12 N(2 s.f.), which is6mgas expected at the minimum condition. - 8
Common trap.
Stopping after the energy step. Writing
½ m v₀² = mg(2L)and solving givesv₀ = √(4gL) = 6.3 m/s, which looks like a complete solution and is a standard distractor. It is the speed that just barely delivers the bob to the top with zero speed — but a string cannot push, so tension goes to zero and the string slackens well before that point, and the bob leaves the circular path. The energy equation never mentions the string, so it cannot detect the failure. Always imposeT ≥ 0at the top as a second, separate condition. - 9
Similar NEET-style question.
A bob of mass
mon a light string of lengthLis whirled in a vertical circle at exactly the minimum speed for the string to stay taut. What is the ratio of the string tension at the lowest point to that at the highest point, and what is the ratio of the bob's speed at the lowest point to that at the highest point? *(Expect the tension ratio to be undefined or infinite, sinceT_top = 0at the minimum condition — a deliberate check that the student knows the top tension vanishes. The speed ratio is√5 : 1.)*
What to remember before solving Motion Vertical Circle questions
Where do students lose marks on Motion Vertical Circle?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
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.
More in Work, Energy and Power: 11 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Motion Vertical Circle questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Motion Vertical Circle?
1 recurring pattern from past papers — click to collapse
Bob of mass m on string of length l given horizontal velocity v₀ at lowest point; find condition for string to remain taut throughout the vertical circle. Minimum: v₀² ≥ 5gl. At top: v_top² = v₀² - 4gl ≥ gl. Energy conservation + tension at any angle.
Multi StepHard
Common distractors
uses only energy not tension
Energy gives speed but tension constraint sets the floor
Sources
NCERT refs: Class 11 Physics Chapter 5, p.82
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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ShortA string cannot push — where the minimum speed at the top really comes from (2:31)
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