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Chapter 5 of 16 · Physics
Circular Motion
Circular Motion averages 2 MCQs per paper, focused on centripetal acceleration, rotational kinematics, and moment of inertia.
Circular Motion is a Physics chapter on the official PMDC MDCAT 2026 syllabus, contributing roughly 2 MCQs to the 36-MCQ Physics section. Mastering the core concepts below typically secures the full chapter weightage.
Angular kinematics
For uniform circular motion of radius r and linear speed v, angular speed ω = v/r, period T = 2π/ω, and frequency f = 1/T. Angular displacement is in radians (1 rev = 2π rad). The angular kinematic equations mirror the linear ones: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ. A wheel accelerating from rest at α = 2 rad/s² for 5 s reaches ω = 10 rad/s and turns θ = 25 rad ≈ 4 revolutions.
Centripetal acceleration and force
A body in circular motion has centripetal acceleration ac = v²/r = ω²r directed toward the centre. The net inward force is Fc = mv²/r = mω²r. A 1000 kg car taking a 50 m radius bend at 20 m/s needs Fc = 1000·400/50 = 8000 N. On a banked road of angle θ with no friction, tan θ = v²/(rg) gives the design speed; for a 50 m bend at 20 m/s with g = 10, tan θ = 400/500 = 0.8, so θ ≈ 38.7°.
Rotational dynamics
Torque τ = r×F = rF sin θ; angular acceleration α = τ/I, where I is the moment of inertia. For a point mass, I = mr². For a uniform disk about its centre, I = ½MR²; solid sphere I = (2/5)MR²; hollow sphere I = (2/3)MR²; thin rod about centre I = ML²/12, about end ML²/3. Rotational KE = ½Iω². A solid sphere rolling without slipping has total KE = ½mv² + ½Iω² = ½mv²(1 + 2/5) = (7/10)mv².
Angular momentum and conservation
L = Iω = mvr (for a particle with v perpendicular to r). When the net external torque is zero, L is conserved — a skater pulling arms in spins faster because I decreases and ω increases proportionally. A planet in elliptical orbit moves faster at perihelion than aphelion, the basis of Kepler's second law; L = mvr remains constant.
Artificial satellites and orbital speed
For a satellite in circular orbit at radius r from Earth's centre, gravitational pull supplies centripetal force: GMm/r² = mv²/r, so v = √(GM/r). At Earth's surface r = R, v ≈ 7.9 km/s (first cosmic speed). Geostationary orbit has T = 86 400 s, giving r ≈ 4.22×10⁷ m or about 36 000 km altitude. The escape velocity ve = √(2GM/R) ≈ 11.2 km/s — exactly √2 times the orbital speed at the surface, a relation HRW emphasises in Chapter 13.
Key Concepts
- Angular velocity
- Centripetal acceleration & force
- Banking of roads
- Orbital motion
- Moment of inertia
Worked MCQs
Q1. A stone tied to a 1 m string is whirled at 4 rad/s. Its centripetal acceleration is:
- A. 4 m/s²
- B. 8 m/s²
- C. 16 m/s² ✓
- D. 32 m/s²
Explanation: a_c = ω²r = 16·1 = 16 m/s².
Common trap: Using ωr = 4 m/s² gives the linear speed, not the acceleration.
Q2. Moment of inertia of a uniform disk of mass M and radius R about its centre is:
- A. MR²
- B. ½MR² ✓
- C. (2/5)MR²
- D. (2/3)MR²
Explanation: Standard result for a uniform disk about its symmetry axis.
Common trap: Confusing with the (2/5)MR² value for a solid sphere.
Q3. An ice-skater pulls in her arms, halving her moment of inertia. Her angular speed:
- A. Halves
- B. Stays the same
- C. Doubles ✓
- D. Quadruples
Explanation: L = Iω is conserved (no external torque). If I → I/2, then ω → 2ω.
Common trap: Thinking KE is conserved — it actually increases as the skater does work pulling her arms in.
Q4. Escape velocity from Earth's surface is approximately:
- A. 7.9 km/s
- B. 9.8 km/s
- C. 11.2 km/s ✓
- D. 30 km/s
Explanation: v_e = √(2GM/R) ≈ 11.2 km/s.
Common trap: Choosing 7.9 km/s — that is the orbital speed at the surface, lower by a factor of √2.
Q5. A solid sphere rolls without slipping at speed v. Fraction of its KE that is translational is:
- A. 1/2
- B. 2/7
- C. 5/7 ✓
- D. 2/5
Explanation: Total KE = (1 + 2/5)·½mv² = (7/10)mv²; translational ½mv² is 5/7 of total.
Common trap: Choosing 2/5 — that is the rotational fraction, not the translational.
Frequently Asked Questions
Is centripetal force a real force?
It is the name for the net inward force, supplied by tension, gravity, friction, or normal reaction depending on the setup. It is not an additional force.
Why does a banked road allow higher speeds without friction?
The horizontal component of the normal force supplies the centripetal force, so friction is no longer required for the design speed.
How is angular momentum different from linear momentum?
Angular momentum L = r×p depends on the choice of origin and is conserved when external torques are zero, just as p is conserved when external forces are zero.
Which rolls faster down an incline, a ring or a solid sphere?
The solid sphere, because it has a smaller moment of inertia coefficient (2/5 vs 1) and so converts more PE to translational KE.
What determines the height of a geostationary orbit?
Matching the orbital period to Earth's rotation (24 h) via T² = 4π²r³/(GM) gives r ≈ 4.22×10⁷ m, or about 36 000 km altitude.
How Circular Motion Is Tested
MDCAT questions on Circular Motion are a mix of recall (definitions, classifications), application (predict outcomes, interpret diagrams), and basic numerical/analytical reasoning. PMDC papers from 2020–2025 emphasized the concepts above; older UHS papers (2008–2019) tested them too, with slight variations in question framing.
Practice
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See the full MDCAT 2026 syllabus or browse all Physics chapters.