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.You need to login to perform this action.
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