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Sticking of Person With Wall of Rotor

NEET > Physics > Laws of Motion > Friction > Sticking of Person With Wall of Rotor

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NEET Physics — Friction

Sticking of Person With Wall of Rotor – Complete Notes, Revision, Important Questions & Downloads

Sticking of Person With Wall of Rotor covers the TOC subtopic Minimum Angular Velocity. A person of mass m stands in contact with the inner wall of a cylindrical drum (radius r) that rotates about its vertical axis. When the floor is removed, the person is held against the wall by friction. The centripetal (centrifugal in non-inertial frame) force provides the normal reaction N = mω²r. Friction = μN = μmω²r must support weight mg. Minimum angular velocity: ω_min = √(g/μr). NEET tests this as a direct formula numerical or conceptual question on rotor rides.

⬇ Download Notes PDFView Important Questions →
Minimum Angular VelocityFriction Ch.5Circular Motion + Friction
Expected QuestionsQ
0–1
NEET asks: (1) find ω_min = √(g/μr) given μ and r, (2) find minimum number of rotations per second (n_min = ω_min/(2π)), (3) conceptual: how does ω_min depend on mass (it doesn't — mass cancels).
Time Required⏱
20 min
10 min: derive ω_min from N=mω²r (centripetal), f=μN, condition f≥mg → μmω²r≥mg → ω≥√(g/μr). 10 min: convert to frequency n=ω/(2π) and period T=2π/ω, practise 3 numericals.
Difficulty⚡
Medium
Requires combining circular motion (centripetal force provides normal reaction) with friction (vertical friction supports weight). The key insight: in a rotating drum, the wall pushes the person inward (centripetal), and this inward push creates the normal force. Friction on the vertical surface (μN) then acts upward.
NRI USA Curriculum GapUS
Medium
AP Physics 1 covers circular motion and friction separately but does not combine them in the 'rotor ride' configuration. The rotor problem is a NEET staple that NRI students need to practise explicitly. The formula ω_min = √(g/μr) should be memorised or quickly derivable.
1Subtopics
4+Practice Questions
4Free Downloads
20 minPrep Time
⬇ Get Free Downloads

NEET Weightage — Sticking of Person With Wall of Rotor

Friction (Chapter 5)
NEET YearQuestions from this TopicBarMarks
20241
 
1 Q
4
20230
 
0 Q
0
20221
 
1 Q
4
20210
 
0 Q
0
20201
 
1 Q
4
20190
 
0 Q
0
6-Year Total (2019–2024)0–2 0–8
SETUP: Person (mass m) inside a cylindrical drum (radius r) rotating about a vertical axis at angular velocity ω. The floor is dropped/removed. Person is held against the wall by friction. FORCES: Normal force N from the drum wall on the person: N = mω²r (centripetal force equation — the net inward force needed for circular motion = mω²r; this is provided entirely by N). Friction from the drum wall on person: f = μN = μmω²r (acting vertically upward, opposing downward sliding). Weight: mg (acting downward).
CONDITION: For person to remain on the wall without sliding down: f ≥ mg → μmω²r ≥ mg → μω²r ≥ g → ω² ≥ g/(μr) → ω ≥ √(g/μr). MINIMUM: ω_min = √(g/μr). Mass m cancels — ω_min is mass-independent. Minimum frequency: n_min = ω_min/(2π) = (1/2π)√(g/μr). Minimum time period: T_max = 2π/ω_min = 2π√(μr/g) [the maximum period for which person doesn't slide].

DEPENDENCE: ω_min increases as μ decreases (less friction → need more centrifugal compression). ω_min decreases as r increases (larger drum → need less ω for same centripetal force). ω_min ∝ 1/√(μr). A heavier person and a lighter person need the same ω_min (mass cancels) — consistent with the analogous result in the block-on-cart problem.
📊
0.5
Avg Questions / Year
🎯
12
Total Marks (6 yrs)
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Direct
Pattern
⚠️
Medium
Difficulty

How to Solve the Rotor Ride Problem

1

Step 1 — Write the centripetal force equation The drum wall pushes the person inward to keep them in circular motion: N = mω²r. This is the normal force from the wall on the person.

2

Step 2 — Write the friction condition Friction (upward) ≥ weight (downward): μN ≥ mg → μmω²r ≥ mg. Cancel m: μω²r ≥ g.

3

Step 3 — Solve for ω_min ω_min = √(g/μr). Convert to frequency: n_min = ω_min/(2π). Convert to velocity: v_min = ω_min × r = r√(g/μr) = √(gr/μ).

Study Materials — Sticking of Person With Wall of Rotor

PDF · Cheat Sheet · MCQ Set · PYQ
📘
Full Notes
Rotor of radius r, angular velocity ω. N=mω²r (centripetal). Friction=μmω²r (upward). Condition: μω²r≥g. ω_min=√(g/μr). n_min=(1/2π)√(g/μr). v_min=√(gr/μ). Mass-independent.
1 subtopicCircular motion + frictionDirect formula
Download Notes
📗
Formula Sheet
ω_min=√(g/μr); n_min=ω_min/(2π); T_max=2π/ω_min=2π√(μr/g); v_min=√(gr/μ).
4 formulas1 pageQuick reference
Download Sheet
📙
MCQ Practice
8 MCQs: compute ω_min, find n_min per second, compare two persons of different mass (same ω_min), vary r or μ.
8 MCQs2 formulasSolved
Download MCQs
📒
PYQ
NEET previous year questions on rotor/centrifuge friction with complete solutions.
3+ year-tagged Qs2015–2024Solved
Download PYQs

Subtopics — Sticking of Person With Wall of Rotor

2-Column Table
Column AColumn B
Minimum Angular Velocity↗

Rapid Revision — Sticking of Person With Wall of Rotor

Concept → Trap → Example

1) Minimum Angular Velocity — Rotor/Drum Problem

Minimum Angular Velocity

PROBLEM: A person (mass m) in a cylindrical rotor (inner radius r). The rotor spins vertically about its central axis with angular velocity ω. The floor is removed. Person presses against the wall and is supported by friction. Find the minimum ω for the person not to slide down. INERTIAL (GROUND) FRAME: Person moves in a horizontal circle of radius r with angular velocity ω. Net horizontal centripetal force (inward) = mω²r. Only horizontal force on person = Normal force N from the drum wall. Therefore N = mω²r (directed toward the axis, centripetal). Vertical forces: friction f (upward, from wall) and weight mg (downward). For person to not slide: f ≥ mg. At limiting (minimum) condition: f = μN. So μN ≥ mg → μmω²r ≥ mg. Cancel m: μω²r ≥ g → ω² ≥ g/(μr) → ω ≥ √(g/μr). RESULT: ω_min = √(g/μr). KEY CONVERSIONS: Minimum frequency (revolutions per second): n_min = ω_min/(2π) = [1/(2π)]√(g/μr). Maximum safe period T (above which person slides): T_max = 1/n_min = 2π/ω_min = 2π√(μr/g). Minimum linear speed (tangential): v_min = ω_min × r = r√(g/μr) = √(gr/μ).

  • MASS INDEPENDENCE: ω_min = √(g/μr) contains no mass term. Both a 50 kg person and a 100 kg person in the same rotor need the same ω_min. This is because raising mass m increases both the required friction needed (mg) and the available friction (μmω²r) proportionally — the m cancels in the condition μmω²r ≥ mg. In amusement park rotor rides, all passengers (regardless of weight) are safe at the same speed.
  • NON-INERTIAL (ROTATING) FRAME: In the frame of the rotor, the person is stationary. Centrifugal force (pseudo-force) = mω²r directed radially outward (away from axis). This presses the person against the wall. Normal force N = mω²r (wall on person, inward) balances the centrifugal force. Friction = μN = μmω²r (upward). Weight = mg (down). Equilibrium: μmω²r = mg → ω = √(g/μr). Both frames give the same ω_min.
  • SPEED FORM: v_min = √(gr/μ). This is the minimum tangential speed of the person for sticking. Note: v = ωr, so ω_min = v_min/r = √(gr/μ)/r = √(g/μr) ✓. Larger radius r: v_min increases (more speed needed for larger drum) but ω_min = v/r actually decreases (less angular velocity). NEET sometimes asks for v_min instead of ω_min — use v_min = √(gr/μ) or equivalently v_min = ω_min × r.
Example (NEET-style)Rotor radius r = 2 m, μ = 0.5 between person and wall. ω_min = √(g/μr) = √(10/(0.5×2)) = √(10/1) = √10 ≈ 3.16 rad/s. n_min = 3.16/(2π) ≈ 0.5 rev/s (30 rpm). v_min = ω_min × r = 3.16 × 2 = 6.32 m/s. Check: N = mω²r = m×10×2 = 20m N. Friction = 0.5×20m = 10m N = mg ✓.

US Curriculum Gaps — Sticking of Person With Wall of Rotor

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

Rotor Ride Problem in AP Physics 1

AP Physics 1 covers uniform circular motion and centripetal force but does not include the 'rotor ride' (person sticking to vertical drum wall) as a standard problem type. NRI students need to identify that in this configuration, the centripetal force equation gives N = mω²r, and vertical friction from this normal force must support the person's weight. The rotor problem is a NEET classic combining two concepts (circular motion + friction) that AP Physics tests separately.

  • AP Physics 1: centripetal force and friction both covered, but not combined in vertical-axis drum configuration
  • NEET: ω_min=√(g/μr) is a direct formula — derive once, memorise, apply in 30 seconds
  • Key: horizontal centripetal equation gives N; vertical friction equation gives the condition on ω

Centrifuge and Circular Motion with Friction in University Physics

Halliday & Resnick includes circular motion problems but the rotor-ride-with-friction configuration is not always a standard textbook example. University Physics students learn to apply Newton's law in circular motion but need NEET-specific practice on the rotor problem. The formula ω_min = √(g/μr) requires recognising that the centripetal acceleration provides the effective gravity for the normal force, which is the unusual conceptual element.

  • US university: Newton's law in circular motion covered; rotor-ride-friction not always a standard example
  • NEET: also tests minimum linear speed v_min=√(gr/μ) and minimum rotational frequency n_min=(1/2π)√(g/μr)
  • Practise with r=1m, μ=0.5: ω_min=√(20)≈4.47 rad/s, n_min≈0.71 rev/s, T_max≈1.4 s

NEET-Style Practice — Sticking of Person With Wall of Rotor

4 Questions
1A person with a mass m stands in contact against the wall of a cylindrical drum (rotor) of radius 2 m. The coefficient of friction between person and wall is 0.2. What is the minimum angular velocity of the drum for the person to be held against the wall when the floor is lowered? (g = 10 m/s²)Minimum Angular Velocity
2.5 rad/s
5 rad/s
10 rad/s
√5 rad/s ≈ 2.24 rad/s
ω_min = √(g/μr) = √(10/(0.2×2)) = √(10/0.4) = √25 = 5 rad/s. Answer: Option B. Verify: N = mω²r = m×25×2 = 50m. Friction = 0.2×50m = 10m = mg ✓. Option A (2.5 rad/s): ω²=6.25; N=m×6.25×2=12.5m; f=0.2×12.5m=2.5m < mg=10m — insufficient. The formula √(g/μr) = √(10/0.4) = √25 = 5 (exactly, no square root needed here).
2A rotor has radius r = 5 m and μ = 0.5 between person and wall. What is the minimum rotational frequency (revolutions per second) for a person to stick to the wall?Minimum Frequency
1/(π) rev/s ≈ 0.318 rev/s
1 rev/s
2/π rev/s
10/π rev/s
ω_min = √(g/μr) = √(10/(0.5×5)) = √(10/2.5) = √4 = 2 rad/s. n_min = ω_min/(2π) = 2/(2π) = 1/π ≈ 0.318 rev/s. Answer: Option A. Check: N=mω²r=m×4×5=20m. Friction=0.5×20m=10m=mg ✓. Again, the formula gives a clean answer because √(g/μr)=√4=2.
3Two persons of masses m₁ = 50 kg and m₂ = 100 kg are in the same rotor (r, μ). Their minimum angular velocities ω₁ and ω₂ satisfy:Mass Independence
ω₁ = ω₂ (mass does not affect ω_min)
ω₂ = 2ω₁ (heavier person needs more ω)
ω₁ = 2ω₂
ω₁/ω₂ = √(m₂/m₁) = √2
ω_min = √(g/μr) — no mass term. Both persons need the same ω_min. Answer: Option A. This is because in the condition μmω²r ≥ mg, the mass m cancels. A heavier person experiences proportionally more centripetal normal force (mω²r) and proportionally more weight (mg); the ratio is 1:1, so the mass doesn't affect the minimum ω. In a rotor amusement park ride, all customers are safe at the same rotational speed.
4If the radius of the rotor is doubled (from r to 2r) while μ stays the same, the minimum angular velocity becomes:Effect of Radius Change
ω_min/√2
ω_min/2
√2 × ω_min
ω_min (unchanged)
ω_min = √(g/μr). If r → 2r: new ω_min = √(g/μ×2r) = √(g/2μr) = ω_min/√2. Answer: Option A. Doubling radius decreases ω_min by factor √2 (≈0.707). Intuitively: larger drum provides more centripetal force at the same ω (N=mω²r increases with r), so less ω is needed for the same friction support. Note: minimum tangential speed v_min = ω_min × r = (ω_min/√2) × 2r = √2 × ω_min × r = √2 × v_min (increases with r).

Practice Problems — Sticking of Person With Wall of Rotor

Click "Reveal Answer" after attempting
1A rotor (r = 1 m, μ = 0.4). Find ω_min, n_min, and v_min. (g = 10 m/s²)
ω=5 rad/s; n=5/(2π)≈0.80 rev/s; v=5 m/s
ω=√25=5; n≈0.80; v=5 m/s (consistent)
ω=2 rad/s; n≈0.32 rev/s; v=2 m/s
ω=10 rad/s; n≈1.59 rev/s; v=10 m/s
👁 Reveal Answer
ω_min=√(g/μr)=√(10/0.4)=√25=5 rad/s. n_min=5/(2π)≈0.796 rev/s. v_min=ω×r=5×1=5 m/s. Answer: Options A and B are consistent (same values). Verify: N=mω²r=m×25×1=25m. Friction=0.4×25m=10m=mg ✓.
2A rotor has μ = 0.25 between person and wall. The floor is removed at 4 rad/s. The minimum radius the rotor must have for the person to stick is:
r = 2.5 m
r = 4 m
r = 10 m
r = 1 m
👁 Reveal Answer
From ω_min=√(g/μr): ω=4, ω²=16. 16=g/(μr)=10/(0.25r) → 16=10/(0.25r) → 16×0.25r=10 → 4r=10 → r=2.5 m. Answer: Option A. At r=2.5 m, ω=4 rad/s is exactly ω_min. For r>2.5 m, the rotor at 4 rad/s would be above ω_min (person sticks). For r<2.5 m, the person slides (ω=4 < ω_min at smaller r).
3A person is in a 3 m radius rotor spinning at 6 rad/s. The floor is suddenly removed. For the person to stay, the minimum μ required is:
μ = 0.5
μ = 0.185
μ = 0.06
μ = 3
👁 Reveal Answer
From ω_min=√(g/μr): ω=6, ω²=36. g/(μr)≤ω² → μ≥g/(ω²r)=10/(36×3)=10/108=5/54≈0.0926≈0.093. Closest option: B (0.185 is too high). Let me recompute: 10/(36×3)=10/108≈0.0926. The minimum μ needed is approximately 0.093 — floor must have at least this μ. None of the options match exactly; but the calculation gives μ_min≈0.093. If options are approximate: B (0.185) = 2×0.093 (wrong), C (0.06) < 0.093 (insufficient). The correct answer is μ_min ≈ 0.093. If the NEET option set had 1/11 ≈ 0.091 or 5/54, that would be correct.
4A rotor of radius 2 m has μ = 0.5 between person and wall. If the angular speed is 3 rad/s, which statement is correct about the friction force on the person?
Friction = μmω²r = 0.5×m×9×2 = 9m N (sliding down at this ω)
Friction = mg = 10m N (exactly balanced — person is at ω_min)
The person slides because ω = 3 < ω_min = √(g/μr)
Friction = mg and person is static — ω = √(10/(0.5×2)) = √10 ≈ 3.16 > 3, so person slides
👁 Reveal Answer
ω_min=√(g/μr)=√(10/(0.5×2))=√(10/1)=√10≈3.16 rad/s. Current ω=3 < ω_min=3.16. At ω=3: N=mω²r=m×9×2=18m. Max friction=μN=0.5×18m=9m N < mg=10m N. Friction cannot support weight — person slides DOWN. Answer: Option C (person slides because ω < ω_min). Option D correctly states that ω_min≈3.16>3 and person slides.

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FAQ — Sticking of Person With Wall of Rotor

Notes · Downloads · Revision · Important Questions
What is the formula for minimum angular velocity in the rotor problem?
ω_min = √(g/μr), where g is gravitational acceleration, μ is the coefficient of static friction between person and wall, and r is the radius of the drum. This is derived from: centripetal force equation N = mω²r; friction condition μN ≥ mg → μmω²r ≥ mg → ω ≥ √(g/μr).
Why is ω_min independent of mass?
In the condition μmω²r ≥ mg, the mass m appears on both sides and cancels: μω²r ≥ g → ω ≥ √(g/μr). Both the centripetal normal force (mω²r) and the weight (mg) are proportional to m. So heavier and lighter people need the same ω_min in the same rotor — this is why rotor amusement rides are safe for all weights at one rotation speed.
What is the minimum tangential speed for the rotor problem?
v_min = ω_min × r = √(g/μr) × r = √(gr/μ). Alternatively, from the condition: N = mv²/r (centripetal) and μN ≥ mg → μmv²/r ≥ mg → v² ≥ gr/μ → v_min = √(gr/μ). Note: v_min increases with r (larger drum needs more speed), while ω_min decreases with r.
How does ω_min change if the radius is quadrupled?
ω_min = √(g/μr). If r → 4r: new ω_min = √(g/4μr) = ω_min/2. Quadrupling radius halves ω_min. In general, ω_min ∝ 1/√r. Doubling r: ω_min → ω_min/√2. The minimum speed v_min = √(gr/μ) changes as: v_min' = √(g×4r/μ) = 2v_min (doubles when r quadruples).
What is the maximum time period for the rotor?
T_max = 2π/ω_min = 2π/√(g/μr) = 2π√(μr/g). Above this period (T > T_max), the rotor rotates too slowly and the person slides down. Below this period (T < T_max), the person is held safely. Note: T_max and ω_min are inversely related — shorter period (faster rotation) is safer.
What physical mechanism holds the person up in a rotor?
The rotation of the drum presses the person against the wall — this creates a horizontal normal force N = mω²r. The wall exerts this normal force on the person's back/shoulders. Friction between the person's clothing (or body) and the drum wall acts vertically upward. This upward friction supports the person's weight mg. The floor is not needed for support — friction from the vertical wall does the job.
Can I use v instead of ω in NEET problems?
Yes. NEET may give v (linear speed) or ω (angular velocity). Convert using v = ωr. The condition becomes: from N = mv²/r (using v for centripetal force): μN ≥ mg → μmv²/r ≥ mg → v² ≥ gr/μ → v_min = √(gr/μ). Both ω_min = √(g/μr) and v_min = √(gr/μ) are correct forms.
What if the friction is kinetic in the rotor problem?
At exactly ω_min, the person is on the verge of sliding — we use limiting (kinetic = static at limit) friction. If ω < ω_min, the person slides down with kinetic friction. In this case friction = μ_k × mω²r < mg, and there is a net downward force causing downward acceleration. For NEET, the problem always asks for the minimum ω to PREVENT sliding, so we use the limiting condition with static friction equal to its maximum value μ × N.
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Minimum Angular Velocity

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