Kinematics of Circular Motion — IISER Physics MCQs with Solutions
Free IISER Physics Kinematics of Circular Motion MCQs with step-by-step solutions (8 questions). Part of Circular Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Kinematics of Circular Motion · easy · theory
In non-uniform circular motion, the total acceleration of a particle is:
A. √(a_c² + a_t²), since the centripetal and tangential accelerations are perpendicular ✓ Correct
B. Only a_c
C. a_c − a_t
D. a_c + a_t
Solution: a_c (radial, v²/r) ⊥ a_t (tangential, αr); they add vectorially: a = √(a_c² + a_t²). Trap: forgetting a_t in non-uniform motion.
Q2 — Kinematics of Circular Motion · medium · theory
A particle moves in a circle with uniformly INCREASING speed. The angle between its total acceleration and its velocity is:
A. Exactly 90°
B. Acute (between 0° and 90°) ✓ Correct
C. Obtuse
D. Zero
Solution: a_t is along v (speeding up) and a_c is ⊥ v; the resultant leans forward of the radius — an acute angle with v. (Obtuse if slowing down; 90° only if uniform.)
Q3 — Kinematics of Circular Motion · easy · numerical
A wheel turns at 240 rpm. Its angular velocity in rad/s is:
A. 240 rad/s
B. 4π rad/s
C. 2π rad/s
D. 8π ≈ 25.1 rad/s ✓ Correct
Solution: ω = 2πN/60 = 2π×240/60 = 8π ≈ 25.1 rad/s.
Q4 — Kinematics of Circular Motion · medium · numerical
A particle moves in a circle of radius 2 m. At an instant its speed is 4 m/s and its speed is increasing at 3 m/s². The magnitude of its total acceleration is:
A. 3 m/s²
B. √73 ≈ 8.5 m/s² ✓ Correct
C. 8 m/s²
D. 11 m/s²
Solution: a_c = v²/r = 8 m/s², a_t = 3 m/s²; a = √(64 + 9) = √73 ≈ 8.5 m/s². Trap: not 8 + 3.
Q5 — Kinematics of Circular Motion · hard · numerical
A particle starts from rest on a circle of radius 1 m with constant angular acceleration α = 2 rad/s². At t = 1 s the magnitude of its TOTAL acceleration is:
A. 4 m/s²
B. √20 ≈ 4.5 m/s² ✓ Correct
C. 2 m/s²
D. 6 m/s²
Solution: ω = αt = 2 rad/s ⇒ a_c = ω²r = 4 m/s²; a_t = αr = 2 m/s²; a = √(16+4) = √20 ≈ 4.5 m/s².
Q6 — Kinematics of Circular Motion · medium · numerical
A wheel accelerates uniformly from rest to 20 rad/s in 4 s. The number of revolutions completed in this time is:
A. 20/π ≈ 6.4 ✓ Correct
B. 80/π
C. 40
D. 10
Solution: θ = ½αt² with α = 5 rad/s²: θ = ½×5×16 = 40 rad ⇒ n = 40/2π = 20/π ≈ 6.4 rev.
Q7 — Kinematics of Circular Motion · hard · theory
For a particle in circular motion, the angular velocity vector ω⃗ and the linear velocity v⃗ satisfy v⃗ = ω⃗ × r⃗. The vector ω⃗ points:
A. Along the rotation axis, perpendicular to the plane of the circle ✓ Correct
B. Along the tangent
C. Opposite to v⃗
D. Radially outward
Solution: ω⃗ is an axial vector ⊥ to the plane (right-hand rule); v⃗ = ω⃗ × r⃗ then correctly gives the tangential direction.
Q8 — Kinematics of Circular Motion · hard · numerical
A particle's angular position is θ = t³ − 3t² (rad). Its angular acceleration is zero at t =
A. 2 s
B. 0.5 s
C. 3 s
D. 1 s ✓ Correct
Solution: ω = 3t² − 6t; α = 6t − 6 = 0 ⇒ t = 1 s (ω is momentarily at its minimum).