NEET Physics — Friction

Motion of Insect in Rough Bowl – Complete Notes, Revision, Important Questions & Downloads

Motion of Insect in Rough Bowl covers the TOC subtopic Maximum Height Formula. An insect crawls up the inner rough spherical bowl (radius r, coefficient of friction μ). It can climb only until the limiting friction balances the tangential (gravity) component along the bowl surface. The maximum height the insect reaches above the bottom is h = r[1 − 1/√(1+μ²)]. NEET tests this as a direct substitution numerical — given r and μ, compute h — or as a conceptual MCQ asking how h depends on μ.

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Maximum Height FormulaFriction Ch.5Limiting Friction on Curved Surface

Expected QuestionsQ

0–1

NEET tests this topic typically as: (1) given r and μ, compute maximum height h = r[1 − 1/√(1+μ²)], (2) find y (the radial height from centre at which insect stays) = r/√(1+μ²), (3) conceptual: for what μ does insect reach the rim? For reaching the rim (h = r), the equation demands 1/√(1+μ²) = 0, which requires μ → ∞ — so insect can never reach the rim for finite μ on a smooth enough bowl.

Time Required⏱

20 min

10 min to derive the formula from free body diagram at the limiting position — resolve forces tangentially and normally at angle θ. 10 min to memorise the result h = r[1 − 1/√(1+μ²)] and practice 3–4 substitution numericals with different r and μ values. The geometry (relating h, r, y using h = r − y, and y = r/√(1+μ²)) is the key setup to understand.

Difficulty⚡

Medium

The formula derivation requires visualising the geometry of the bowl: at height h, the insect is at a point where the radius makes angle θ with the vertical. The normal to the bowl at that point is along the radius (inward). The weight mg has components normal to bowl (mgcosθ) and tangential along the bowl (mgsinθ). Limiting friction (μN = μmgcosθ) must balance the tangential component (mgsinθ), giving tanθ = μ → sinθ/cosθ = μ. Then: y = rcosθ = r/√(1+μ²), h = r − y.

NRI USA Curriculum GapUS

Medium

AP Physics 1 does not cover insect-in-bowl geometry as a named problem. The physics — friction on a curved surface at limiting position — is extension beyond standard AP curriculum. NRI students who are strong in geometry can derive the result but typically haven't seen it before. The key insight that the normal to a spherical bowl passes through the centre (i.e., normal along radius) is essential and should be memorised.

1Subtopics

4+Practice Questions

4Free Downloads

20 minPrep Time

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NEET Weightage — Motion of Insect in Rough Bowl

Friction (Chapter 5)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	0	0 Q	0
2022	1	1 Q	4
2021	0	0 Q	0
2020	1	1 Q	4
2019	0	0 Q	0
6-Year Total (2019–2024)	0–2		0–8

GEOMETRY SETUP: Bowl of radius r. Insect at limiting position makes radius-to-vertical angle θ. Vertical depth from top of bowl to insect position: y = rcosθ (measured from bowl centre downward). Height climbed from bottom: h = r − y = r(1 − cosθ). Normal force N on insect at angle θ: N = mgcosθ (component of weight along the radius, inward toward centre). Tangential force: mgsinθ (component along surface, directed toward bottom). Limiting friction = μN = μmgcosθ.
DERIVATION OF h: At limiting position, friction balances tangential gravity: μN = mgsinθ → μmgcosθ = mgsinθ → μ = tanθ. From tanθ = μ: using right triangle, sinθ = μ/√(1+μ²), cosθ = 1/√(1+μ²). Therefore y = rcosθ = r/√(1+μ²). Height h = r − y = r[1 − 1/√(1+μ²)] = r[√(1+μ²) − 1]/√(1+μ²). This is the maximum height the insect can reach.
KEY RESULT: h = r[1 − 1/√(1+μ²)]. Special cases: μ → 0 (smooth bowl): h → 0 (insect cannot climb). μ → ∞ (very rough): 1/√(1+μ²) → 0 → h → r (insect can almost reach the rim but theoretically needs infinite friction to reach h = r exactly). For μ = 1 (tanθ = 1 → θ = 45°): h = r[1 − 1/√2] = r[1 − 0.707] = 0.293r. As μ increases, h increases monotonically — more friction allows climbing higher.

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0.3

Avg Questions / Year

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Total Marks (6 yrs)

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Direct

Pattern

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Medium

Difficulty

How to Solve the Insect Bowl Problem for NEET

Step 1 — Draw FBD at the limiting position (angle θ) Draw the circular bowl cross-section. Mark the insect at angle θ from the vertical (angle the radius makes with vertical). Three forces on insect: weight mg (downward), normal force N (along radius, toward centre), friction f (tangential, directed toward the bottom of bowl — friction opposes the tendency to slide down, so acts upward along the surface).

Step 2 — Resolve along normal and tangential directions NORMAL equilibrium: N = mgcosθ (weight component along radius balances normal). TANGENTIAL equilibrium at limiting position: f = mgsinθ → μN = mgsinθ → μmgcosθ = mgsinθ → tanθ = μ. This gives θ = arctan(μ). From tanθ = μ: cosθ = 1/√(1+μ²).

Step 3 — Compute h from the geometry y = r cosθ = r/√(1+μ²) (vertical distance from bowl centre to insect, measured downward). Height climbed from bottom h = r − y = r − r/√(1+μ²) = r[1 − 1/√(1+μ²)]. Trap: do not confuse y (the depth of insect below the bowl centre level) with h (height above the bowl's bottom).

Study Materials — Motion of Insect in Rough Bowl

PDF · Cheat Sheet · MCQ Set · PYQ

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Full Notes

Setup: insect at angle θ from vertical in bowl of radius r. Normal force N = mgcosθ. Tangential force = mgsinθ. Limiting: μmgcosθ = mgsinθ → tanθ = μ → cosθ = 1/√(1+μ²). y = r/√(1+μ²). Maximum height h = r[1 − 1/√(1+μ²)]. Interpretation: h increases with μ; h → r as μ → ∞.

1 subtopicLimiting equilibriumGeometry derivation

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Formula Sheet

tanθ = μ; cosθ = 1/√(1+μ²); y = r/√(1+μ²); h = r[1−1/√(1+μ²)]. Check: μ=0→h=0; μ=1→h=r(1−1/√2)=0.293r; μ→∞→h→r.

4 formulas1 pageQuick reference

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MCQ Practice

8 MCQs: compute h for various r and μ; find μ given h and r; conceptual questions on h dependence on μ; limiting cases.

8 MCQs1 formulaSolved

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PYQ

Previous year NEET questions on insect-in-bowl and similar curved-surface friction problems with complete solutions.

2+ year-tagged Qs2015–2024Solved

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Subtopics — Motion of Insect in Rough Bowl

2-Column Table

Column A	Column B
Maximum Height Formula↗

Rapid Revision — Motion of Insect in Rough Bowl

Concept → Trap → Example

1) Maximum Height Formula — Insect in Bowl Derivation

Maximum Height Formula

SETUP: Hemispherical bowl of radius r, rough inner surface (coefficient μ). Insect of mass m crawls from the bottom. At the highest reachable point, the insect is on the verge of sliding down — friction is at its limiting value. GEOMETRY: At limiting point, the radius to the insect makes angle θ with the vertical. Let O = centre of the spherical bowl, A = insect's position. The vertical distance from O to A (measured downward) = y = r cosθ. Height of insect above the bottom (= the nadir of the bowl) = h = r − y = r(1 − cosθ). FORCES at A: (1) Weight mg: vertically downward. (2) Normal force N: directed from A toward the centre O (i.e., along the radius, inward). On a spherical bowl, the surface normal at any point always passes through the geometric centre. (3) Friction f: along the surface, directed upward (insect tends to slide down, so friction acts up the bowl wall). EQUATIONS: Along normal (radius direction): N = mg cosθ [weight component toward centre balances normal]. Along tangential (surface direction): f = mg sinθ [weight component along surface, downhill]. AT LIMITING CONDITION: f = μN → mg sinθ = μ(mg cosθ) → tan θ = μ. From tan θ = μ: using Pythagoras, in right triangle with opposite=μ, adjacent=1, hypotenuse=√(1+μ²): sinθ = μ/√(1+μ²), cosθ = 1/√(1+μ²). RESULT: y = r cosθ = r/√(1+μ²). h = r − y = r[1 − 1/√(1+μ²)].

GEOMETRIC SHORTCUT: Instead of finding θ first, use the relation y = r/√(1+μ²) directly. Start with tanθ = μ → this defines a right triangle (sides 1, μ, √(1+μ²)). cosθ = 1/√(1+μ²). y = r cosθ = r/√(1+μ²). Then h = r − y = r − r/√(1+μ²) = r[√(1+μ²) − 1]/√(1+μ²). For NEET: the height formula h = r[1 − 1/√(1+μ²)] is the single result to memorise.
DEPENDENCE ON μ: h increases as μ increases. For μ = 0 (frictionless bowl): h = 0 (insect stays at bottom, any attempt to climb fails immediately). For μ = 1: h = r(1 − 1/√2) ≈ 0.293r. For μ = √3: cosθ = 1/√(1+3) = 0.5 → θ = 60°; h = r(1 − 0.5) = 0.5r. For μ → ∞: h → r (insect can reach the rim). Maximum h is limited to r (the rim height = r above the bottom for a hemispherical bowl).
COMMON NEET TRAP: Students often confuse y (the vertical coordinate of the insect measured from the bowl's geometric centre) with h (height above the bottom). The bowl's bottom is at vertical distance r BELOW the centre. So h = r − y (where y is the depth below centre, not the height above bottom). Verify: at bottom (θ = 90°): y = r cos90° = 0, h = r − 0 = r? No: at bottom θ = 0 (radius points straight down), y = r cos0 = r, h = r − r = 0 ✓. At rim (θ = 90°): y = r cos90° = 0, h = r ✓.

Example (NEET-style)A hemispherical bowl of radius 20 cm has μ = 0.5 between insect and bowl. Maximum height the insect can reach: h = r[1 − 1/√(1+μ²)] = 20[1 − 1/√(1+0.25)] = 20[1 − 1/√1.25] = 20[1 − 1/1.118] = 20[1 − 0.894] = 20×0.106 = 2.12 cm. Verification: tanθ = 0.5 → θ = 26.57°; cosθ = 0.894; y = 20×0.894 = 17.88 cm; h = 20 − 17.88 = 2.12 cm ✓.

US Curriculum Gaps — Motion of Insect in Rough Bowl

Topics in this section are in NEET but may be framed differently in US physics courses.

Curved Surface Friction in AP Physics 1

AP Physics 1 covers friction on flat inclined planes but not on curved concave surfaces. The concept that the normal to a spherical surface always points along the radius is not explicitly practised in AP Physics 1. NRI students who understand flat-surface friction can solve this by treating the bowl surface at angle θ locally as a flat incline — the same limiting equilibrium applies, but the geometry to find cosθ = 1/√(1+μ²) requires an additional step not present in AP problems.

AP Physics 1: friction on flat inclines covered; curved surface (bowl) friction not in standard curriculum
NEET: the insect-in-bowl problem is a named application of limiting equilibrium on a curved surface
Key insight: normal to spherical bowl passes through centre — this is the geometry fact needed to resolve forces

Limiting Equilibrium on Concave Surfaces in University Physics

Halliday & Resnick (University Physics) covers equilibrium of objects on curved surfaces in statics chapters, but not the specific insect-bowl problem as a named result. The formula h = r[1−1/√(1+μ²)] does not appear in standard US textbooks. NRI students should be comfortable with the technique: (1) identify the point where friction reaches its maximum, (2) set up force balance in normal and tangential directions, (3) solve for the geometric parameters.

US university physics: equilibrium on curved surfaces covered in statics but without the insect-bowl application
NEET: requires memorisation of h = r[1−1/√(1+μ²)] or rapid derivation from tanθ=μ
Geometry shortcut: from tanθ=μ, write cosθ=1/√(1+μ²) directly — saves steps in exam conditions

NEET-Style Practice Questions — Motion of Insect in Rough Bowl

4 Questions

1A hemispherical bowl of radius 40 cm has a coefficient of friction μ = 0.75 between the bowl surface and an insect. What is the maximum height the insect can reach above the bottom of the bowl?Maximum Height Formula

10 cm

8 cm

20 cm

24 cm

Using h = r[1 − 1/√(1+μ²)] = 40[1 − 1/√(1+0.5625)] = 40[1 − 1/√1.5625] = 40[1 − 1/1.25] = 40[1 − 0.8] = 40×0.2 = 8 cm. Option B (8 cm) is correct. Check: 1/√1.5625 = 1/1.25 = 0.8 because √(1.5625) = 1.25 (since 1.25² = 1.5625 ✓). Alternative: tanθ = μ = 0.75 = 3/4 → in right triangle (3,4,5 scaled): sinθ = 3/5, cosθ = 4/5. y = rcosθ = 40×(4/5) = 32 cm. h = 40 − 32 = 8 cm ✓. The 3-4-5 triangle recognition makes this solvable without calculator.

2An insect reaches a maximum height of h = r/2 in a hemispherical bowl of radius r. What is the coefficient of friction μ between insect and bowl?Finding μ

μ = 1/√3

μ = 1/2

μ = √3

μ = 1/√2

From h = r[1 − 1/√(1+μ²)] = r/2: 1 − 1/√(1+μ²) = 1/2 → 1/√(1+μ²) = 1/2 → √(1+μ²) = 2 → 1+μ² = 4 → μ² = 3 → μ = √3. Wait — that gives μ = √3 (Option C). Let me re-derive: h = r/2 → r[1−1/√(1+μ²)] = r/2 → 1 − 1/√(1+μ²) = 0.5 → 1/√(1+μ²) = 0.5 → √(1+μ²) = 2 → μ = √3. So Option C (μ=√3) is actually correct. Verify: tanθ = μ = √3 → θ = 60°. cosθ = 0.5. y = r×0.5 = r/2. h = r − r/2 = r/2 ✓. Option A (1/√3) gives h = r[1−1/√(1+1/3)] = r[1−1/√(4/3)] = r[1−√3/2] ≈ r[1−0.866] = 0.134r ≠ r/2.

3Two hemispherical bowls have the same radius r but different friction coefficients μ₁ = 0.6 and μ₂ = 0.8. An identical insect crawls in each bowl. The ratio h₁/h₂ of heights reached is:Comparing Heights

0.6/0.8 = 0.75

[1−1/√1.36]/[1−1/√1.64]

0.8/0.6 = 4/3

1 (same height)

h₁ = r[1−1/√(1+0.36)] = r[1−1/√1.36] = r[1−0.8575] = r×0.1425. h₂ = r[1−1/√(1+0.64)] = r[1−1/√1.64] = r[1−0.7809] = r×0.2191. h₁/h₂ = 0.1425/0.2191 ≈ 0.65. Exact form: h₁/h₂ = [1−1/√1.36]/[1−1/√1.64]. Option B gives this exact expression. Option A (0.75) incorrectly scales h proportional to μ (h is NOT linear in μ). Option D (same height) is wrong — h increases with μ. The formula h = r[1−1/√(1+μ²)] is concave in μ, so doubling μ does not double h.

4At the highest point an insect can reach in a rough hemispherical bowl, the normal force exerted by the bowl on the insect equals:Forces at Limiting Height

mg/√(1+μ²)

μmg

mg×μ/√(1+μ²)

At limiting position: N = mgcosθ. Since tanθ = μ, cosθ = 1/√(1+μ²). Therefore N = mg/√(1+μ²). Option B is correct. Verify with limiting friction: f = μN = μmg/√(1+μ²) = mg×μ/√(1+μ²). And tangential force = mgsinθ = mg×μ/√(1+μ²) ✓ (friction balances tangential gravity at limiting position). Option A (mg) would mean θ=0 (bottom of bowl, no friction needed). Option C (μmg) confuses friction force with normal force. The NEET key formula: N at limiting height = mg/√(1+μ²), not mg.

Practice Problems — Motion of Insect in Rough Bowl

Click "Reveal Answer" after attempting

1A hemispherical bowl of radius 10 cm. Coefficient of friction between insect and bowl = 1/√3 (≈0.577). Find (a) the angle θ at the limiting position, (b) the maximum height h above the bowl's bottom.

(a) θ = 30°; (b) h = 10(1−√3/2) = 10(1−0.866) = 1.34 cm

(a) θ = 30°; (b) h = 10(1−1/2) = 5 cm

(a) θ = 60°; (b) h = 10(1−0.5) = 5 cm

(a) θ = 45°; (b) h = 10(1−1/√2) = 2.93 cm

👁 Reveal Answer

tanθ = μ = 1/√3 → θ = 30°. cosθ = cos30° = √3/2 ≈ 0.866. y = r cosθ = 10 × (√3/2) = 5√3 ≈ 8.66 cm. h = r − y = 10 − 5√3 = 10(1 − √3/2) ≈ 10×0.134 = 1.34 cm. Answer: Option A. Check via formula: h = r[1−1/√(1+μ²)] = 10[1−1/√(1+1/3)] = 10[1−1/√(4/3)] = 10[1−√3/2] = 10×0.134 = 1.34 cm ✓.

2For the insect-in-bowl problem, derive an expression for the friction force on the insect at the limiting position in terms of mg and μ.

f = μmg (independent of angle)

f = mg×μ/√(1+μ²) — friction at limiting position

f = mg×sinθ where θ = arccos(μ)

f = μmg/cosθ

👁 Reveal Answer

At limiting position: N = mgcosθ = mg/√(1+μ²) [since cosθ = 1/√(1+μ²)]. Friction f = μN = μ×mg/√(1+μ²) = mg×μ/√(1+μ²). This equals the tangential gravity force: mgsinθ = mg×μ/√(1+μ²) [since sinθ = μ/√(1+μ²)] ✓. Note: for μ=1 (θ=45°): f = mg/√2 ≈ 0.707mg. Answer: f = mgμ/√(1+μ²). This is also the weight component along the bowl surface, confirming balance. Answer: Option B.

3A spherical bowl of radius R = 25 cm. If μ = 3/4, find the maximum height h the insect can reach. Use the 3-4-5 Pythagorean triple to simplify.

h = 25(1 − 4/5) = 25×0.2 = 5 cm

h = 25×(3/5) = 15 cm

h = 25×(1/5) = 5 cm both give same

h = 25×(4/5) = 20 cm

👁 Reveal Answer

μ = 3/4. From tanθ = 3/4, recognise the 3-4-5 right triangle: opposite=3, adjacent=4, hypotenuse=5. cosθ = 4/5. y = R cosθ = 25×(4/5) = 20 cm. h = R − y = 25 − 20 = 5 cm. Alternatively: h = R[1−1/√(1+9/16)] = 25[1−1/√(25/16)] = 25[1−4/5] = 25×0.2 = 5 cm. Answer: Option A and C both give h = 5 cm (consistent). The 3-4-5 trick: whenever μ = 3/4, cosθ = 4/5, and h = R×(1−4/5) = R/5.

4In the insect-bowl problem, what is the minimum coefficient of friction required for the insect to reach exactly half the radius height (h = r/2)?

μ = 1

μ = √3

μ = 1/√3

μ = 2

👁 Reveal Answer

Set h = r/2: r[1−1/√(1+μ²)] = r/2 → 1−1/√(1+μ²) = 1/2 → 1/√(1+μ²) = 1/2 → √(1+μ²) = 2 → 1+μ² = 4 → μ² = 3 → μ = √3 ≈ 1.73. Answer: Option B (μ = √3). Verification: tanθ = √3 → θ = 60°. cosθ = 0.5. y = 0.5r. h = r − 0.5r = 0.5r = r/2 ✓. Note: reaching h = r/2 requires μ = √3 ≈ 1.73 (quite rough). For comparison, μ = 1 gives h = r(1−1/√2) ≈ 0.293r (less than half).

Physics — Friction Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Maximum Height Formula: Write tanθ = μ and cosθ = 1/√(1+μ²) from memory. Then derive y = r/√(1+μ²) and h = r−y = r[1−1/√(1+μ²)] without looking at notes.

Maximum Height Formula: Verify the limiting condition: mgsinθ = μmgcosθ → tanθ = μ. Draw FBD at the limiting position showing N along the radius (toward centre), weight mg downward, friction f tangentially upward along the bowl.

Maximum Height Formula: Practise recognising Pythagorean triples from μ: μ=3/4→cosθ=4/5 (3-4-5 triple); μ=√3→cosθ=1/2 (1-√3-2 triple); μ=1→cosθ=1/√2. These speed up calculation in MCQ exams.

Maximum Height Formula: Check special cases: μ=0→h=0; μ=1→h=r[1−1/√2]=0.293r; μ=√3→h=r/2; μ→∞→h→r. Sketch h vs μ graph: monotonically increasing, asymptoting to r.

Maximum Height Formula: Practise the reverse problem: given h and r, find μ. From h=r[1−1/√(1+μ²)]: h/r=1−1/√(1+μ²) → 1/√(1+μ²)=(1−h/r) → 1+μ²=1/(1−h/r)² → μ=√[1/(1−h/r)²−1]. Solve numerically with h=0.293r: μ=1.

FAQ — Motion of Insect in Rough Bowl

Notes · Downloads · Revision · Important Questions

What is the maximum height formula for an insect in a rough hemispherical bowl?

h = r[1 − 1/√(1+μ²)], where r is the bowl's radius and μ is the coefficient of static friction between insect and bowl surface. This is derived by setting limiting friction = tangential gravity at the highest reachable point: μmgcosθ = mgsinθ → tanθ = μ → cosθ = 1/√(1+μ²) → y = rcosθ = r/√(1+μ²) → h = r − y.

Why does the insect stop climbing at angle θ = arctan(μ)?

At angle θ from the vertical, the gravity component along the bowl surface (tangential) = mgsinθ (pulling the insect downward along the bowl). The friction force (opposing downward slide) = μN = μmgcosθ. The insect can climb as long as friction ≥ tangential gravity: μmgcosθ ≥ mgsinθ → μ ≥ tanθ. The highest point where friction just balances tangential gravity is at tanθ = μ. Above this, tanθ > μ, so gravity exceeds friction and the insect slides down.

Why is the normal force N = mgcosθ and not mg?

At angle θ from the vertical, the weight mg acts downward. The bowl's normal force acts along the radius (toward the centre). The component of weight along the radius (perpendicular to the bowl surface) = mgcosθ. This is the component the bowl pushes against. The remaining component = mgsinθ is tangential (along the bowl surface, directed toward the bottom). So N = mgcosθ, not mg.

How does h change as μ increases?

h = r[1−1/√(1+μ²)] increases monotonically as μ increases. For μ=0: h=0 (frictionless, insect cannot climb). For μ=1: h≈0.293r. For μ=√3: h=r/2. For μ→∞: h→r (rim). The relationship is NOT linear — doubling μ does not double h. h approaches r asymptotically, meaning infinite friction is needed to reach the rim exactly.

What is the significance of the Pythagorean triple trick for this problem?

When μ = m/n is a ratio that forms a Pythagorean triple with another integer, the calculation simplifies greatly. Example: μ=3/4 → tanθ=3/4 → (3,4,5) triple → cosθ=4/5. Then h = r(1−4/5) = r/5. Example: μ=4/3 → (3,4,5) → cosθ=3/5 → h=r(1−3/5)=2r/5. These integer triangles avoid square roots in the final answer.

What happens if the bowl is not hemispherical but a full sphere?

For a complete sphere, the insect is on the INSIDE of the bowl. The geometry is the same: normal force is still along the radius (toward centre), and h = r[1−1/√(1+μ²)] still applies for the height climbed from the bottom. The formula holds regardless of whether the bowl is a hemisphere or a full sphere, as long as the insect is on the inside curved surface.

Can the insect reach the rim of the bowl?

No, for any finite μ. At the rim, h = r, which requires 1−1/√(1+μ²) = 1 → 1/√(1+μ²) = 0 → μ = ∞. For any physically realizable friction coefficient (finite μ), the insect reaches a maximum height h < r. As μ increases, h approaches r asymptotically but never equals r.

What is the normal force on the insect at the limiting position?

N = mgcosθ = mg/√(1+μ²), using cosθ = 1/√(1+μ²). For μ=1: N = mg/√2 ≈ 0.707mg. For μ=√3: N = mg/2 = 0.5mg. Note: N < mg at the limiting height — part of the weight is supported by friction (the tangential component). At the bowl's bottom (h=0, θ=0): N = mg (full weight on normal force). As insect climbs higher (larger θ), normal force decreases.

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Motion of Insect in Rough Bowl

Motion of Insect in Rough Bowl – Complete Notes, Revision, Important Questions & Downloads

1) Maximum Height Formula — Insect in Bowl Derivation

Curved Surface Friction in AP Physics 1

Limiting Equilibrium on Concave Surfaces in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Maximum Height Formula

Subtopics

Maximum Height Formula

NEET > Physics > Laws of Motion Chapters

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Motion of Insect in Rough Bowl

Overview content

Motion of Insect in Rough Bowl – Complete Notes, Revision, Important Questions & Downloads

1) Maximum Height Formula — Insect in Bowl Derivation

Curved Surface Friction in AP Physics 1

Limiting Equilibrium on Concave Surfaces in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Maximum Height Formula

Subtopics

Maximum Height Formula

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

Comments (0)