Angular momentum (left or right movement)

AI Thread Summary
The discussion focuses on a physics problem involving a rectangular sign swinging and being struck by a snowball. The sign has a mass of 2.40 kg and swings with a maximum angular displacement of 25.0°. When the snowball, weighing 400 g and traveling at 160 cm/s, strikes the sign, the conservation of angular momentum is applied to determine the system's movement. Calculations reveal that the angular momentum of the sign is greater than that of the snowball, indicating that the system will move to the left. The conversation emphasizes the importance of understanding angular momentum and moment of inertia in solving such problems.
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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