question_answer

A particle falls freely near the surface of the earth Consider a fixed point O (not vertically below the particle) on the ground. Then pickup the incorrect alternative or alternatives.

A)  The magnitude of angular momentum of the particle about O is increasing

B)  The magnitude of torque of the gravitational force on the particle about O is decreasing

C)  The moment of inertia of the particle about O is decreasing

D)  The magnitude of angular velocity of the particle about O is increasing

Correct Answer: B

Solution :

 If student will use angular momentum = mvr. He/she may conclude answer [A] as r is decreasing angular momentum must decrease hence [A] is incorrect. The magnitude of angular momentum of particle about = mvd Since speed v of particle increases, its angular momentum about O increases. Magnitude of torque of gravitational force about O= mgd\[\Rightarrow \]constant Moment of inertia of particle about \[O=m{{r}^{2}}\] Hence Ml of particle about 0 decreases. Angular velocity of particle about \[O=\frac{v\sin \theta }{r}\] \[\because \] v and sin \[\theta \] increases and r decreases. \[\therefore \]angular velocity of particle about O increases.