Angular momentum in projectile motion

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In projectile motion, angular momentum is not conserved due to the influence of external forces, specifically gravity, which acts as an external torque. While it may seem that angular momentum could be conserved since the gravitational force acts along the direction of displacement, this is misleading because the projectile is not an isolated system. The gravitational field is treated as uniform and vertical, complicating the conservation of momentum. Analyzing a projectile from different perspectives, such as an observer on the ground, reveals that both linear and angular momentum change as the projectile falls. Therefore, in typical projectile motion scenarios, angular momentum is not conserved.
Yatin
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Homework Statement



Is angular momentum of a projectile in projectile motion conserved ?[/B]

Homework Equations

The Attempt at a Solution


Gravitional force vector is acting along direction of displacement vector. Thus external torque is 0. Thus angular momentum is conserved throughout projectile motion. Am i right?

I am confused because I read somewhere that it's not.
 
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Momentum (linear or angular) is conserved in isolated systems. That is, where no external forces or torques are applied to elements of the system, those quantities are conserved.

In typical projectile motion problems near the surface of the Earth the gravitational force is effectively an external force acting on the projectile and we don't consider the motion of the Earth that results from the gravitational interaction. Further, the gravitational field is taken to be uniform and oriented vertically, not radially.

So the projectile does not comprise an isolated system in such cases, and momentum (of either sort) is not conserved in such a "system".

Why don't you set yourself a task to clear your confusion? Suppose you have a projectile which is dropped from rest from some height (pick a height). An observer at ground level and situated some distance away (pick a distance) calculates the linear and angular momentum of the projectile (from his point of view) at its initial position and just before it hits the ground. What will he find?
 
Yatin said:
Gravitional force vector is acting along direction of displacement vector.

This is not necessarily true. If the projectile is moving horizontally, the displacement will have a horizontal component which means the gravitational force will not be parallel to it
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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