Angular momentum (left or right movement)

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Homework Help Overview

The problem involves a thin uniform rectangular sign that swings vertically and is impacted by a snowball. The context is angular momentum and its conservation during the collision between the sign and the snowball.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the moment of inertia of the sign and its angular momentum when passing the vertical. There are attempts to calculate angular velocity and compare angular momenta of the sign and the snowball.

Discussion Status

Participants have engaged in calculations related to angular momentum and have provided feedback on each other's work. There is a clear progression in the discussion, with some guidance offered on finding angular velocity and comparing momenta.

Contextual Notes

The original poster's question about the direction of movement after the collision remains open, with various calculations being explored without a definitive conclusion on the system's motion.

careman
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Homework Statement



A thin uniform rectangular sign hangs vertically above the
door of a shop. The sign is hinged to a stationary horizontal
rod along its top edge. The mass of the sign is 2.40 kg
and its vertical dimension is 50.0 cm. The sign is swinging
without friction, becoming a tempting target for children
armed with snowballs. The maximum angular displacement
of the sign is 25.0° on both sides of the vertical. At a
moment when the sign is vertical and moving to the left, a
snowball of mass 400 g, traveling horizontally with a velocity
of 160 cm/s to the right, strikes perpendicularly the
lower edge of the sign and sticks there.

Is the system (sign-snowball) going to move left or right?

Homework Equations



Conservation of angular momentum.

The Attempt at a Solution



Lost
 
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What's the angular momentum of the sign when it passes the vertical? Hint: Start by finding its moment of inertia.
 
Thank you for your reply.

The moment of inertia is I=\frac{1}{3}ML^2, therefore the angular momentum of the sign when it passes the vertical is L_s=Iω=(\frac{1}{3}ML^2)ω.
 
careman said:
The moment of inertia is I=\frac{1}{3}ML^2, therefore the angular momentum of the sign when it passes the vertical is L_s=Iω=(\frac{1}{3}ML^2)ω.
Excellent. Now find ω.

Once you've done that, figure out the initial angular momentum of the snowball.
 
I found the angular velocity ω from the conservation of energy:

\frac{1}{2}Iω^2=Mgh
ω=\sqrt{\frac{3g(1-cosθ)}{L}}=2.35 rad/s

The angular momentum of the sign is L_s=\frac{1}{3}ML^2=0.47 Nms while the angular momentum of the snowball is L_b=mvL=0.32 Nms

Comparing these two, the angular momentum of the sign is greater than the angular momentum of the ball, therefore the system is going to move to the left.
 
Good work!
 
Thank you for your help and for checking my work. I really appreciate it.
 

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