Rolling Motion (Rolling without Slipping)
Rotation
55
JEE Qs
25%
Hard
100
min
Master the concept of the Instantaneous Axis of Rotation (IAR) and understand when to apply energy conservation versus dynamic force/torque equations for efficient problem-solving.
🧮 Key Formulas
✅ Key Points for JEE
- 1Rolling without slipping is a combination of translational motion of the center of mass and rotational motion about the center of mass, governed by v_CM = Rω and a_CM = Rα.
- 2The point of contact with the ground is the Instantaneous Axis of Rotation (IAR). Calculating torque, angular momentum, or kinetic energy about the IAR can often simplify problem-solving significantly.
- 3For pure rolling, static friction acts at the contact point. This friction does NO work, meaning mechanical energy (translational KE + rotational KE + PE) is conserved if no other non-conservative forces are present.
- 4To analyze dynamics, draw a free-body diagram, apply Newton's second law for translation (ΣF = ma_CM) and rotation (Στ = Iα) about either the CM or the IAR, along with the no-slip condition.
- 5The kinetic energy of a rolling body can be written as (1/2)I_IARω^2, where I_IAR is the moment of inertia about the instantaneous axis of rotation (contact point).
⚠️ Common Mistakes
- ✕Assuming friction always opposes the direction of CM motion, instead of opposing the tendency of relative motion at the contact point.
- ✕Incorrectly applying the rolling without slipping condition (v_CM = Rω or a_CM = Rα) when slipping *is* occurring.
- ✕Assuming friction does work in pure rolling, leading to incorrect application of energy conservation principles.
- ✕Not selecting the appropriate axis for torque calculations (e.g., using CM when IAR simplifies things, or vice-versa).
- ✕Using an incorrect moment of inertia (e.g., I_CM instead of I_P when calculating angular momentum about the contact point).
📝 Practice Questions
See allQ28.A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in 2025 (22 Jan Shift 1) JEE Main Previous Year Paper the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. (1) 32 7 MR2 (2) 329 MR2 (3) 17 32 MR2 (4) 1332 MR2
Q48.The position vectors of two 1 kg particles, (A) and (B), are given by→rA = (α1t2^i + α2t^j + α3t^k)m , respectively; →rB = (β1t^i + β2t2^j + β3t^k)m (α1 = 1 m/s2, α2 = 3nm/s, α3 = 2 m/s, β1 = 2 m/s, β2 = −1 m/s2, β3 = 4pm/s), where t is time, n and p −→ are constants. At t = 1 s, VA = →VB and velocities →VA and →VB of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is √Lkgm2 s−1 . The value of L is _______ .
Q28.The torque due to the force (2^i + ^j + 2^k) about the origin, acting on a particle whose position vector is (^i + ^j + ^k), would be (1) ^i −^k (2) ^i + ^k (3) ^j + ^k (4) ^i −^j + ^k
Q49.A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is in SI unit. The value of α is __________. √FαM
Q45.A solid sphere of mass ' m ' and radius ' r ' is allowed to roll without slipping from the highest point of an inclined plane of length ' L ' and makes an angle 30∘ with the horizontal. The speed of the particle at the bottom of the plane is v1 . If the angle of inclination is increased to 45∘ while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2 . The ratio v21 : v22 is (1) 1 : √2 (2) 1 : √3 (3) 1 : 3 (4) 1 : 2 2025 (23 Jan Shift 1) JEE Main Previous Year Paper
Q40.A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination 45∘ . If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be (1) 1 g (2) 1 g √2 3√2 (3) √2 g (4) √2g 3
NCERT Chapters
- Class 11 Physics Ch 7: System of Particles and Rotational Motion