Angular Momentum of a shop sign

In summary: Yes, the position vector is x=0 y=50cm. To find the angular momentum of the sign, you would need to find its rotational velocity.
  • #1
lastguymade
8
0

Homework Statement



Hi guys! I'm having trouble with this Problem. 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 2.40kg and its vertical dimension is 50.0cm. 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 degrees on both sides of the vertical. At a moment when the sign is vertical and moving to the left, a snowball of mass 400g, traveling horizontally with a velocity of 160cm/s to the right strikes perpendicularly the lower edge of the sign and sticks there. (a) Calculate the angular speed of the sign immediately before the impact. (b) Calculate its angular speed immediately after the impact. (c) The spattered sign will swing up through what maximum angle?


Homework Equations



for part (a) I'm not sure if I'm supposed to use conservation of mechanical energy or if I should do the vector product of the sign at the maximum angle (since its swinging without friction) multiplied by the vector of the sign at the origin.

for part (b) I know momentum is conserved but I really don't understand how linear momentum converts to angular momentum or vice versa. I am also confused as to how get the moment of inertias for both the sign and snowball before impact and the combined moment of inertia after.

for part (c) I'm completely lost



The Attempt at a Solution



Ok so for (a) if I use conservation of kinetic energy I get mgh=.5mw^2

if I take height as h=.5cos25 I get w=2.98

If I do it the second way I mentioned I get w=L/I = 1.892

(I got the moment of Inertia by looking on this forum where someone mentioned it but I don't know how they got it.)

for (b) I got the angular momentum of the sign by multiplying I=.2 *w (2.98)=.596 (This would be wrong if my first calculation was wrong) and then I added it with the momentum of the ball to get the total momentum which is conserved after to get L(total)=-.044

I then divided that number by the new I (.2+.1) and got w=-.146

(c) I don't know at all.

I'm sorry guys, I'm really confused by angular momentum but I figured this would be the best place to look. If my basic understanding is too flawed then just tell me and I'll go back to the books but any tips would be appreciated.

Cheers!
 
Physics news on Phys.org
  • #2
Hi lastguymade, welcome to PF!

Kinetic energy is not conserved in this situation, because it is a completely inelastic collision (the snowballs hits the sign and then sticks there).

EDIT: In hindsight, I see that although you called it conservation of kinetic energy, what you really meant was conservation of mechanical energy. Okay, I will re-evaluate your solution. But it's still an inelastic collision, so not all of the kinetic energy of the snowball has to become potential energy. Some of it can be lost (as sound, heat, work done to deform and break pieces off the snowball, etc).
 
  • #3
Hello and thanks for the reply!

Yes I did mean conservation of mechanical energy, my bad. I was applying it to the first question though (angular speed of the sign immediately before impact) so the snowball doesn't have anything to do with that I think?
 
  • #4
I think that the way to solve this problem is to apply conservation of angular momentum about the pivot point of the sign. The snowball has some angular momentum about this point just before the collision, and so does the sign. You can equate the sum of those to the angular momentum after the collision, keeping in mind that your moment of inertia has changed.

Moments of inertia can just be looked up in tables. See for example:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia
 
  • #5
lastguymade said:
Hello and thanks for the reply!

Yes I did mean conservation of mechanical energy, my bad. I was applying it to the first question though (angular speed of the sign immediately before impact) so the snowball doesn't have anything to do with that I think?

Uhh, yeah good point. So, your method is right. I think the problem is that you are mixing up speed (v) and angular speed (ω).

Conservation of energy says:

½mv2 = mgh

Allowing you to solve for v. Then you need to figure out ω and I in order to compute the angular momentum (L). What is the relationship between v and ω?
 
  • #6
Ok so I guess I'm confused as to how I find the angular momentum of the snowball before the collision if I'm not given a position vector. Unless the position vector is x=0 y=50cm?
Also how do I find the angular momentum of the sign without the rotational velocity (which is what I'm trying to get?) Sorry if I'm asking easy questions, I just can't seem to wrap my head around this concept.

Edit: Just read your next post let me see if that helps.
 
  • #7
Ok so solving for v I got 2.98m/s and since v=rw I got w=5.96. Thats the answer to A and I know to find b I apply the conservation of angular momentum. But for the life of me I can't figure out I for the sign or the snowball. Looking at that list I can't find any that fit the scenario with only M and h because those are the only 2 quantities given.
 
  • #8
lastguymade said:
Ok so I guess I'm confused as to how I find the angular momentum of the snowball before the collision if I'm not given a position vector. Unless the position vector is x=0 y=50cm?

Yes exactly. Remember the equation for the angular momentum of a point mass:

L = r x p = r x mv

(where I use boldface to signify vectors). Let's say the point around which you're computing the angular momentum is labelled point Q. In this equation, r is just a vector between point Q and the object. And you're right that it is entirely in the vertical direction in this case. In contrast, the velocity vector (and hence the momentum vector) is entirely horizontal. So the angle between r and p is 90 degrees, and the sine of 90 degrees is 1. Therefore, the magnitude of the angular momentum just becomes:

|L| = rmvsinθ = rmv

You have to be careful to take the direction into account using the right hand rule. In this case, when taking the cross product, in order to rotate from r to p, you have to rotate clockwise, which means that L points "into the page". (The angular momentum of the sign is in the opposite direction, since it is rotating counterclockwise).

lastguymade said:
Ok so solving for v I got 2.98m/s and since v=rw I got w=5.96. Thats the answer to A and I know to find b I apply the conservation of angular momentum. But for the life of me I can't figure out I for the sign or the snowball. Looking at that list I can't find any that fit the scenario with only M and h because those are the only 2 quantities given.

Be careful. There is another wrinkle to part (a) that I did not consider. When we say that the gravitational potential energy is mgh, we are talking about a point mass. In this case, we are dealing with an extended body (it takes up some non-zero amount of space). But no matter -- we can use a point in that body called the centre of mass, which has the special property that all of the mass in that body can be considered to reside there, at that point. So, when you apply the conservation of mechanical energy, h is the height change of the centre of mass, and the speed v that you compute is the speed of the centre of mass. This has implications for how you compute omega.

As for the moment of inertia, the object in question is a thin rectangle. So thin, in fact, that it can be considered to be "one-dimensional." So, if you take the equation in that list for the moment of inertia of a rectangle, and set the width w, to 0, you'll get what you need. (You'll notice that, in that case, the formula reduces to the formula for the moment of inertia of a thin rod, where only the length is important).

EDIT: But it's not pivoting about its centre, it's pivoting around its end, so use the equation for I for a thin rod rotating around one of its ends.
 
  • #9
Wow thank you very much! I see now how the other guys got the moment of inertia and I think I have it figured out. Will post once I get my marks back to see if I got it right.

Thank you once again!
 

1. What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of an object. It is defined as the product of an object's moment of inertia and its angular velocity.

2. How is angular momentum calculated?

The formula for calculating angular momentum is L = Iω, where L represents angular momentum, I represents moment of inertia, and ω represents angular velocity.

3. How does a shop sign's design affect its angular momentum?

The design of a shop sign can affect its angular momentum in several ways. A sign with a larger surface area or more mass will have a higher moment of inertia, resulting in a higher angular momentum. Additionally, a sign with a more aerodynamic design will have a lower drag force, allowing it to maintain its angular velocity for a longer period of time.

4. Can the angular momentum of a shop sign be changed?

Yes, the angular momentum of a shop sign can be changed. This can be accomplished by altering the sign's moment of inertia or angular velocity. For example, changing the shape or weight distribution of the sign can change its moment of inertia, while applying a force can change its angular velocity.

5. How does friction affect the angular momentum of a shop sign?

Friction can have a significant impact on the angular momentum of a shop sign. Friction acts in the opposite direction of an object's motion, causing it to slow down and lose angular momentum over time. This is why it is important to minimize friction when designing a shop sign to ensure it maintains its rotational motion for as long as possible.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
819
  • Introductory Physics Homework Help
2
Replies
45
Views
2K
  • Introductory Physics Homework Help
10
Replies
335
Views
7K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Replies
13
Views
819
  • Introductory Physics Homework Help
Replies
9
Views
891
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
612
Back
Top