Level 1 — RecognitionRotational Mechanics

Rotational Mechanics

20 minutes30 marksprintable — key stays hidden on paper

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 vv and angular velocity ω\omega for a point at distance rr from the axis is: (a) v=ω/rv = \omega / r (b) v=rωv = r\omega (c) v=rω2v = r\omega^2 (d) v=r2ωv = r^2\omega

Q2. The moment of inertia of a solid sphere of mass MM and radius RR about its diameter is: (a) 12MR2\frac{1}{2}MR^2 (b) 23MR2\frac{2}{3}MR^2 (c) 25MR2\frac{2}{5}MR^2 (d) MR2MR^2

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) rF\vec{r}\cdot\vec{F} (b) r×F\vec{r}\times\vec{F} (c) F×r\vec{F}\times\vec{r} giving same sense as (b) (d) F/r\vec{F}/\vec{r}

Q5. The rotational kinetic energy of a body is: (a) 12Iω\frac{1}{2}I\omega (b) Iω2I\omega^2 (c) 12Iω2\frac{1}{2}I\omega^2 (d) 12mv2\frac{1}{2}mv^2

Q6. For rolling without slipping of a wheel of radius RR, the condition connecting centre velocity vv and ω\omega is: (a) v=ω/Rv = \omega/R (b) v=Rωv = R\omega (c) v=R2ωv = R^2\omega (d) v=2Rωv = 2R\omega

Q7. The parallel axis theorem is stated as: (a) I=ICMMd2I = I_{CM} - Md^2 (b) I=ICM+Md2I = I_{CM} + Md^2 (c) I=ICM+MdI = I_{CM} + Md (d) Iz=Ix+IyI_z = I_x + I_y

Q8. Angular momentum for a body rotating about a fixed axis is: (a) L=I/ωL = I/\omega (b) L=IωL = I\omega (c) L=12Iω2L = \frac{1}{2}I\omega^2 (d) L=r×pL = r\times p only

Q9. Angular momentum is conserved when the net external torque is: (a) maximum (b) constant and non-zero (c) zero (d) equal to IαI\alpha

Q10. The perpendicular axis theorem Iz=Ix+IyI_z = I_x + I_y 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. 23MR2\frac{2}{3}MR^2
Q. Ring about central axis 2. 12MR2\frac{1}{2}MR^2
R. Solid disk about central axis 3. 112ML2\frac{1}{12}ML^2
S. Hollow sphere about diameter 4. MR2MR^2

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 ac=rαa_c = r\alpha."

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, τ=dL/dt\vec{\tau} = d\vec{L}/dt."

Q18. "A rigid body is in complete equilibrium if the net force on it is zero."

Q19. "In the parallel axis theorem, dd 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 12mv2+12Iω2\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2."

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (b) v=rωv = r\omega — tangential speed scales linearly with radius. (1)

Q2. (c) 25MR2\frac{2}{5}MR^2 — 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) τ=r×F\vec{\tau} = \vec{r}\times\vec{F} — cross product; magnitude rFsinθrF\sin\theta. (1)

Q5. (c) 12Iω2\frac{1}{2}I\omega^2 — rotational analogue of 12mv2\frac{1}{2}mv^2. (1)

Q6. (b) v=Rωv = R\omega — contact point instantaneously at rest. (1)

Q7. (b) I=ICM+Md2I = I_{CM} + Md^2. (1)

Q8. (b) L=IωL = I\omega for fixed axis. (1)

Q9. (c) zero — no external torque ⇒ LL constant. (1)

Q10. (b) planar (laminar) bodies — theorem derived assuming mass lies in one plane (z=0z=0). (1)

Section B (1 mark each)

Q11. P → 3 (112ML2\frac{1}{12}ML^2, rod about centre) (1)
Q12. Q → 4 (MR2MR^2, ring) (1)
Q13. R → 2 (12MR2\frac{1}{2}MR^2, disk) (1)
Q14. S → 1 (23MR2\frac{2}{3}MR^2, hollow sphere) (1)

Section C (1 verdict + 1 reason)

Q15. FALSE. (1) Centripetal (radial) acceleration is ac=rω2=v2/ra_c = r\omega^2 = v^2/r; rαr\alpha is the tangential acceleration ata_t. (1)

Q16. TRUE. (1) Acceleration on incline a=gsinθ1+I/mR2a = \dfrac{g\sin\theta}{1 + I/mR^2}. Solid sphere I/mR2=2/5=0.4I/mR^2 = 2/5 = 0.4; hollow cylinder I/mR2=1I/mR^2 = 1. Smaller factor ⇒ larger aa ⇒ 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 F=0\sum\vec{F}=0 (translational) and τ=0\sum\vec{\tau}=0 (rotational); zero force alone allows a net couple to rotate the body. (1)

Q19. TRUE. (1) dd is the perpendicular distance between the CM axis and the chosen parallel axis; I=ICM+Md2I = I_{CM}+Md^2. (1)

Q20. TRUE. (1) Total KE = translational KE of CM (12mv2\frac{1}{2}mv^2) + rotational KE about CM (12Iω2\frac{1}{2}I\omega^2). (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"}
]