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Chapter 1: Q35E (page 331)

A2.00kgrock has a horizontal velocity of magnitude 12.0m/swhen it is at point P in fig E10.35. (a) At this instant, what are the magnitude and direction of its angular momentum relative to point O? (b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?


Short Answer

Expert verified

(a) The angular momentum is 115kgm2/s.

(b) The rate of change angular momentum is 125kgm2/s.

Step by step solution

01

Given in the question.

The mass of rock is 2kg.

The horizontal velocity is v=12.0m/s.

02

Formula of angular momentum.

The equivalent of momentum in rotatory motion is known as angular momentum. It is the product of the moment of inertia and angular velocity of the rotating body.

L=mvrcosθ

Here, L is the angular momentum, mis the mass,vis the horizontal velocity.

03

(a) The magnitude and direction of angular momentum relative to point O.

Take the angle as:

θ=180-36.9∘∘=143.1o

Use the formula for angular momentum,

L=mvrsinθ=2kg×12m/s×8 msin143.1∘=115kgm2/s

Thus, the angular momentum is 115kgm2/s.

04

(b) The rate of change of angular momentum at this instant.

Apply formula of angular momentum and solve as follows:

L=mvrcosθ=2kg×9.8m/s2×8mcos36.9∘=125kgm2/s

Thus, the rate of change angular momentum is 125kgm2/s.

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