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⚛️ Physics  ·  System of Particles and Rotational Motion  ·  NEET & JEE

A ball is thrown with backspin onto a rough floor. Initially v (forward) and ωR > v. The friction force on the ball is:

Answer: Forward (in direction of motion).

  • A Forward (in direction of motion)
  • B Backward according to most researchers
  • C Zero in the majority of cases studied
  • D Depends on the surface as widely reported

Correct answer: A. Forward (in direction of motion)

Explanation: Since ωR > v, the contact point moves backward relative to the floor. Kinetic friction acts forward on the ball (opposing relative slip), accelerating it and decelerating its spin until rolling condition v = ωR is met.

axis OPr⊥Fττ = r⊥ × F = r × F × sinθ

Torque about axis O equals the force F multiplied by the perpendicular distance r⊥ from the axis to the line of action of the force, producing rotation.

Concept context

Moment of inertia, torque, angular momentum, rolling motion. Very high JEE weightage.

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