Rolling Motion — Rolling without slipping, KE of rolling
Rotation
12
JEE Qs
8%
Hard
75
min
Master the condition for rolling without slipping (v_cm = Rω) and its implications for kinetic energy and the role of friction.
🧮 Key Formulas
✅ Key Points for JEE
- 1Rolling without slipping implies that the point of contact between the rolling body and the surface is instantaneously at rest relative to the surface.
- 2The condition v_cm = Rω is fundamental for pure rolling and directly relates the translational velocity of the center of mass to the angular velocity.
- 3The total kinetic energy of a rolling body is the sum of its translational kinetic energy (associated with the center of mass motion) and its rotational kinetic energy (associated with rotation about the center of mass).
- 4Static friction is the force responsible for initiating and maintaining pure rolling motion; its direction depends on the tendency of slipping and can act either forwards or backwards.
- 5For objects rolling on inclined planes, the principle of conservation of mechanical energy often provides the most efficient solution, relating initial potential energy to final kinetic energy (both translational and rotational).
⚠️ Common Mistakes
- ✕Incorrectly applying the condition v_cm = Rω when the object is slipping, or not applying it when it is pure rolling.
- ✕Forgetting to include both the translational and rotational components of kinetic energy when using energy conservation principles.
- ✕Misinterpreting the direction of the static friction force; it always opposes the tendency of relative motion at the point of contact, not necessarily the direction of motion of the center of mass.
📝 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