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PhysicsMediumClass 11

Pseudo Force — Non-inertial frames

Laws of Motion

9

JEE Qs

8%

Hard

75

min

Always clearly define your frame of reference (inertial or non-inertial) before applying Newton's laws and pseudo forces to avoid conceptual errors.

🧮 Key Formulas

F_pseudo = -m * a_frame_inertial
F_net_real + F_pseudo = m * a_object_relative_to_frame
F_centrifugal = m * (ω^2) * r

✅ Key Points for JEE

  • 1Pseudo force is an imaginary (fictitious) force introduced only when analyzing motion from a non-inertial (accelerating) frame of reference to make Newton's second law (F=ma) valid.
  • 2The pseudo force on an object acts in a direction opposite to the acceleration of the non-inertial frame relative to an inertial frame, with magnitude F_pseudo = (mass of object) * (acceleration of non-inertial frame).
  • 3When solving problems, choose one frame of reference (either inertial or non-inertial) and consistently apply Newton's laws accordingly; do not mix approaches.
  • 4Centrifugal force is a specific type of pseudo force that acts radially outwards in a uniformly rotating non-inertial frame, with magnitude mω²r.

⚠️ Common Mistakes

  • Applying pseudo forces while working in an inertial frame of reference, leading to incorrect net forces.
  • Incorrectly determining the direction of the pseudo force (it's always opposite to the frame's acceleration relative to an inertial frame).
  • Forgetting to include the mass of the object experiencing the force when calculating the magnitude of the pseudo force.
  • Confusing centrifugal force (pseudo force) with centripetal force (real force) or applying both simultaneously in a rotating frame for the same object.

📝 Practice Questions

See all

Q41.A balloon and its content having mass M is moving up with an acceleration ' a '. The mass that must be released from the content so that the balloon starts moving up with an acceleration ' 3a′ will be (Take ' g ' as acceleration due to gravity) (1) 2Ma (2) 3Ma 3a+g 2a−g (3) 3Ma (4) 2Ma 2a+g 3a−g

2025·MCQMedium

Q33.A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e. dmdt ∝√v . If P is the power delivered to run the belt at constant speed then which of the following relationship is true? 2025 (29 Jan Shift 2) JEE Main Previous Year Paper (1) P ∝√v (2) P ∝v (3) P2 ∝v5 (4) P2 ∝v3

2025·MCQHard

Q29.A car of mass ' m ' moves on a banked road having radius ' r ' and banking angle θ. To avoid slipping from banked road, the maximum permissible speed of the car is v0 . The coefficient of friction μ between the wheels of the car and the banked road is θ θ tan tan 0−rg (1) μ = v2o+rg (2) μ = v2 θ tan θ rg−v2 tan rg+v2o 0 θ θ tan v2o+rg tan 0−rg (4) μ = (3) μ = v2 θ tan θ rg+v2 tan rg−v2o 0

2025·MCQHard

Q47. A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes ( π3 ) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is … … ML.

2025·NumericalMedium

Q5. A wooden block, initially at rest on the ground, is pushed by a force which increases linearly with time t. Which of the following curve best describes acceleration of the block with time: (1) (2) (3) (4)

2024·Graph basedMedium

Q21.Two forces ¯F1 and ¯F2 are acting on a body. One force has magnitude thrice that of the other force and the 1 ) resultant of the two forces is equal to the force of larger magnitude. The angle between →F1 and →F2 is cos−1 ( n . The value of |n| is _____.

2024·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 5: Laws of Motion