Angular momentum (left or right movement)

[careman]

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

 

[Doc Al]

What's the angular momentum of the sign when it passes the vertical? Hint: Start by finding its moment of inertia.

 

[careman]

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 $L_s=Iω=(\frac{1}{3}ML^2)ω$.

 

[Doc Al]

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.

 

[careman]

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.

 

[Doc Al]

Good work!

 

[careman]

Thank you for your help and for checking my work. I really appreciate it.