Rotational Mechanics
Level 1: Recognition (MCQ + Matching + True/False with Justification)
Time Limit: 20 minutes
Total Marks: 30
Section A — Multiple Choice (1 mark each) [10 marks]
Choose the correct option.
Q1. The relation between linear velocity and angular velocity for a point at distance from the axis is: (a) (b) (c) (d)
Q2. The moment of inertia of a solid sphere of mass and radius about its diameter is: (a) (b) (c) (d)
Q3. A rigid body free in space has how many degrees of freedom? (a) 3 (b) 4 (c) 6 (d) 9
Q4. Torque is defined as: (a) (b) (c) giving same sense as (b) (d)
Q5. The rotational kinetic energy of a body is: (a) (b) (c) (d)
Q6. For rolling without slipping of a wheel of radius , the condition connecting centre velocity and is: (a) (b) (c) (d)
Q7. The parallel axis theorem is stated as: (a) (b) (c) (d)
Q8. Angular momentum for a body rotating about a fixed axis is: (a) (b) (c) (d) only
Q9. Angular momentum is conserved when the net external torque is: (a) maximum (b) constant and non-zero (c) zero (d) equal to
Q10. The perpendicular axis theorem applies only to: (a) any 3D body (b) planar (laminar) bodies (c) spheres (d) rods
Section B — Matching (1 mark each pair) [8 marks]
Match each body/axis in Column I with its moment of inertia in Column II.
| Column I | Column II |
|---|---|
| P. Thin rod about centre | 1. |
| Q. Ring about central axis | 2. |
| R. Solid disk about central axis | 3. |
| S. Hollow sphere about diameter | 4. |
Q11. P → ?
Q12. Q → ?
Q13. R → ?
Q14. S → ?
Section C — True/False WITH Justification (2 marks each: 1 verdict + 1 reason) [12 marks]
Q15. "The centripetal acceleration of a point on a rotating body is ."
Q16. "For a rolling body descending an incline, a solid sphere reaches the bottom faster than a hollow cylinder (same mass, radius, height)."
Q17. "The torque equals the rate of change of angular momentum, ."
Q18. "A rigid body is in complete equilibrium if the net force on it is zero."
Q19. "In the parallel axis theorem, is the distance between the axis through the centre of mass and the parallel axis."
Q20. "The total kinetic energy of a rolling body is ."
Answer keyMark scheme & solutions
Section A (1 mark each)
Q1. (b) — tangential speed scales linearly with radius. (1)
Q2. (c) — standard result for solid sphere about diameter. (1)
Q3. (c) 6 — three translational + three rotational degrees of freedom for a free rigid body. (1)
Q4. (b) — cross product; magnitude . (1)
Q5. (c) — rotational analogue of . (1)
Q6. (b) — contact point instantaneously at rest. (1)
Q7. (b) . (1)
Q8. (b) for fixed axis. (1)
Q9. (c) zero — no external torque ⇒ constant. (1)
Q10. (b) planar (laminar) bodies — theorem derived assuming mass lies in one plane (). (1)
Section B (1 mark each)
Q11. P → 3 (, rod about centre) (1)
Q12. Q → 4 (, ring) (1)
Q13. R → 2 (, disk) (1)
Q14. S → 1 (, hollow sphere) (1)
Section C (1 verdict + 1 reason)
Q15. FALSE. (1) Centripetal (radial) acceleration is ; is the tangential acceleration . (1)
Q16. TRUE. (1) Acceleration on incline . Solid sphere ; hollow cylinder . Smaller factor ⇒ larger ⇒ sphere is faster. (1)
Q17. TRUE. (1) Newton's second law for rotation: net external torque equals the time rate of change of angular momentum. (1)
Q18. FALSE. (1) Full equilibrium requires both (translational) and (rotational); zero force alone allows a net couple to rotate the body. (1)
Q19. TRUE. (1) is the perpendicular distance between the CM axis and the chosen parallel axis; . (1)
Q20. TRUE. (1) Total KE = translational KE of CM () + rotational KE about CM (). (1)
[
{"claim":"Solid sphere I/mR^2 = 2/5",
"code":"I=Rational(2,5); result = (I == Rational(2,5))"},
{"claim":"Sphere acceleration exceeds hollow cylinder on incline",
"code":"g,th=symbols('g theta',positive=True); a_sph=g*sin(th)/(1+Rational(2,5)); a_cyl=g*sin(th)/(1+1); result = simplify(a_sph - a_cyl) > 0 if False else bool((a_sph.subs({g:9.8,th:0.5}) - a_cyl.subs({g:9.8,th:0.5}))>0)"},
{"claim":"Rod about centre = 1/12 M L^2",
"code":"M,L=symbols('M L',positive=True); I=integrate((M/L)*x**2,(x,-L/2,L/2)); result = simplify(I - M*L**2/12)==0"},
{"claim":"Parallel axis: disk about edge = 3/2 M R^2",
"code":"M,R=symbols('M R',positive=True); Icm=M*R**2/2; I=Icm+M*R**2; result = simplify(I - Rational(3,2)*M*R**2)==0"}
]