Determining the torque and angular momentum of a bullet due to gravity

AI Thread Summary
The discussion focuses on calculating the torque and angular momentum of a bullet fired from a cannon in the yz plane. The torque is expressed as a function of time, with the textbook providing a formula of -v0mgtcosθ for part A. The user struggles with determining the radius from the origin and the force acting perpendicular to it, leading to confusion in their calculations. For part B, the user seeks clarification on incorporating momentum into the angular momentum formula. The conversation emphasizes the importance of understanding the relationship between torque, angular momentum, and the forces acting on the bullet.
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Homework Statement



A bullet of mass "m" is fired from a cannon located at the origin. The shell moves in the yz plane, where "z" is the vertical coordinate, with an initial velocity of magnitude "v0" at an angle θ above the y-axis.
A) What is the torque on the shell, about the origin, as a function of time?
B) What is the angular momentum of the shell about the origin as a function of time? (Use the definition of angular momentum to solve)

Homework Equations



Torque = r x F = r*F*sinθ
Angular momentum = r x p = r x m*v = r*p*sinθ

The Attempt at a Solution



My textbook says the answer for A) is -v0mgtcosθ (in the \vec{i} direction) and -(1/2)v0mgt2cosθ (\vec{i})

I'm primarily having trouble determining the radius from the origin. I want to say it's v0cosθ*t, and I want to say the force acting perpendicular to the radius is "mg", but then my answer becomes -v0mgtcosθsinθ ... Can someone please help?
Also, I'm not sure how to plug momentum into the angular momentum formula to properly find that for part B.

Any help would be extremely appreciated.
 
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sin(theta) is always 1, because mg acts perpendicular to r.
 
curtains forever said:
sin(theta) is always 1, because mg acts perpendicular to r.
:welcome:

Note that this homework is from 9 years ago.
 
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