Torque — τ = r×F, rotational equilibrium
Rotation
12
JEE Qs
8%
Hard
75
min
Always choose the pivot point strategically to minimize the number of unknown forces contributing to torque calculations, thereby simplifying the problem.
🧮 Key Formulas
✅ Key Points for JEE
- 1Torque is a vector quantity; its direction is given by the right-hand rule (perpendicular to both r and F).
- 2For a body to be in complete equilibrium, both translational (ΣF=0) and rotational (Στ=0) equilibrium conditions must be satisfied simultaneously.
- 3The choice of pivot point for calculating net torque is arbitrary for a body in equilibrium; however, strategically choosing the pivot can simplify calculations by eliminating torques due to unknown forces (e.g., choosing the pivot at the point of application of an unknown reaction force).
- 4The component of the force parallel to the position vector 'r' produces zero torque. Only the perpendicular component of the force (or the perpendicular distance, 'lever arm') contributes to torque.
- 5A force whose line of action passes through the chosen pivot point produces zero torque about that pivot.
⚠️ Common Mistakes
- ✕Incorrectly determining the direction of torque using the right-hand rule, especially for 3D problems.
- ✕Confusing the position vector 'r' with the perpendicular distance (lever arm) and using them interchangeably without proper understanding.
- ✕Forgetting to include all forces (e.g., normal forces, friction, weight, hinge forces) when drawing free-body diagrams and calculating torques.
- ✕Not applying both translational and rotational equilibrium conditions when a problem requires complete static equilibrium analysis of a rigid body.
📝 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