Angular momentum (left or right movement)

  • Thread starter careman
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  • #1
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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
 

Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
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Thank you for your reply.

The moment of inertia is [itex]I=\frac{1}{3}ML^2[/itex], therefore the angular momentum of the sign when it passes the vertical is [itex]L_s=Iω=(\frac{1}{3}ML^2)ω[/itex].
 
  • #4
Doc Al
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The moment of inertia is [itex]I=\frac{1}{3}ML^2[/itex], therefore the angular momentum of the sign when it passes the vertical is [itex]L_s=Iω=(\frac{1}{3}ML^2)ω[/itex].
Excellent. Now find ω.

Once you've done that, figure out the initial angular momentum of the snowball.
 
  • #5
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I found the angular velocity ω from the conservation of energy:

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

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

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.
 
  • #6
Doc Al
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Good work!
 
  • #7
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Thank you for your help and for checking my work. I really appreciate it.
 

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