# A rod losing a support

1. Dec 7, 2013

1. The problem statement, all variables and given/known data

A uniform thin rod of weight W is supported horizontally by two bricks at its
ends. At t=0 one of these bricks is kicked out quickly. Calculate the force on the
other support immediately thereafter - it should be in terms of W.

2. Relevant equations

T=IA F=ma W=mg I=mr^2/3 Linear accel=AL

3. The attempt at a solution

I had attempted to look for the total netforce of the rod. To do this I thought the rod would move down in a sort of pivot of the other support. Using the equations above and some substituting I got the total netforce on the rod was 2W. However, I'm not sure if this is the force that is on the remaining support. Any clarification would be appreciated.

Last edited: Dec 8, 2013
2. Dec 7, 2013

### voko

You could consider the angular acceleration/torque equation where the torque is computed about the supported end of the rod, then the same equation where the torque is computed about some other point, for example, its center of mass.

3. Dec 7, 2013

### Simon Bridge

The rod has not had a chance to start moving yet - but it certainly wants to rotate about the remaining brick.

... Force is a vector ... which direction and where does it act?
We really need to see your working.
(Also - the "equations above" do not include a lower-case "w" ... I take it that's a typo?)

Have you tried drawing a free body diagram for the rod?
What are the individual forces (where do they come from) and where do they act?

4. Dec 8, 2013

Yes that was a typo, I have corrected it. The only forces that I can think of is the force of gravity on the pole and the force of the support acting on the pole. First I had L as the length of the pole. I then used I=mr^2 which became I=m(L/2)^2. I then got T=WL/2 from T=IA because gravity produces a force relative to the pivot point. I plugged this in A=T/I which ended up with A=W2/(Lm). The linear acceleration of the pole is AL and I put that into the F=ma equation leading to F=mAL and then I plug in A=w2/(Lm) which cancels things out to 2W.

Last edited: Dec 8, 2013
5. Dec 8, 2013

### Simon Bridge

That is not the moment of inertia of a rod rotating about one end.
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

So A is the angular acceleration...
That would the the tangential acceleration of the tip of the pole farthest from the pivot ...
The F in F=ma is the force through the center of mass m ... producing linear acceleration a of the entire body.

The tangential acceleration of a point in a rotating object distance r from the pivot is $a=r\tau /I$ ... and you have to have the right I.

Have you tried using voko's suggestion in post #2

Last edited: Dec 8, 2013
6. Dec 8, 2013

My apologies for messing up a few times in what I said. I had put linear accleration as AL/2 which I got under the circular motion section. http://en.wikipedia.org/wiki/Tangential_acceleration#Tangential_and_centripetal_acceleration

I had then replaced the a from F=ma with linear acceleration being a=AL/2. Then using I=[m(L/2)^2]/2 this time and end up with A being 6W/(LM). Then I plug that into F=mAL/2 to get 3W as my new final answer.

I'm not entirely sure where that equation for tangential acceleration comes from because I've never seen it with the r.

I havn't tried voko's suggestion because if I were to calculate the torque around the support wouldn't that cause it to be 0 because of the length. I'm unsure how I would start with his suggestion. I am calculating this with regards to the center of the rod at the moment though (or at least so I think)

7. Dec 9, 2013

### Simon Bridge

That's not how you calculate the moment of inertia.
You need to revise that part of your course notes. For this situation you can look it up:
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

You've seen: $$\tau=I \alpha,\; a=r\alpha$$ ... then do some algebra.

No. If you set up that situation - would it not accelerate? Therefore the torque cannot be zero.

Did you not draw a free-body diagram?

Oh I see... I thought you were doing this for rotation about the pivot?

Lets restart.

You should have a free-body diagram of just the rod.
There are two forces on it - what are they and where do they act?

Voko is suggesting you take two cases - one force causes a torque about the center of mass, the other causes a torque about the pivot point.

8. Dec 9, 2013

### PhanthomJay

great!
good
No, but you had it right the first time when you listed your relevant equations! Stick to your guns.
Yes!
correct your value for I and note that W =mg
look at the tangential acceleration at its COM, not at the end
Correct your F_net = ma equation . F_net consists of 2 forces, one known, the other unknown. Solve for the unknown reaction force.

9. Dec 10, 2013

I can think of two forces after drawing it and one is the force of gravity acting upon it pushing it down and the other is the force of the support pushing it up. May I ask how finding the net force on the rod would help? I'm only looking for the force on support. I don't know the net force itself so finding the other one doesn't help me does it?

Taking a look at that the way that worked it worked out the same as a=A(L/2) ending me with 3W unless I messed up.

10. Dec 10, 2013

### Simon Bridge

If I keep going after what PhanthomJay wrote, I'll just confuse you more.
I'm going to wait a bit.

11. Dec 10, 2013

### ehild

The net force determines the linear acceleration of the CM. The linear acceleration of the CM is equal to the angular acceleration, multiplied by L/2 , if you consider rotation about the brick as pivot.

ehild

12. Dec 10, 2013

### PhanthomJay

Sure it does, the other one is the force of the support on the rod, which is what you are trying to find. The net force in the vertical direction consists of 2 forces, the weight acting down, and the support force acting up. Solve for it using$F_{net}= ma_{com}$, where $a_{com}$ is the tangential acceleration at the com, but please, you must first correctly determine the mass moment of inertia of a rod rotating about one end, which is ___?___? Until you get that right and do the algebra right, you'll be getting wrong answers for angular acceleration, tangential acceleration, etc.

13. Dec 10, 2013

Ok so I did find the correct I being I=mr^2/3 which I changed into I=m(L/2)^2/3 to take into account that I'm calculating it at the center of mass. I had fixed my A=T/I equation to make it into A=6W/(Lm). Then I plugged it into the $F_{net}= ma_{com}$ equation where a is the tangential acceleration of the com. Since tangential acceleration is a=Ar, F=ma becomes F=mA(L/2) for the center of mass. Then I plug in and get 3W.

I feel like I'm missing something obvious here and its not clicking in my head. Where would I go from here then? If there are two forces then wouldn't $F_{net}= ma_{com}$ have something else too?

14. Dec 10, 2013

So I also realized that intertia doesnt really take into account of center of mass and only applies to the whole rod (according to my tutor) so I is actually equal to mL^2/3. I redid the calculations and got 3W/4.

My tutor also mentioned that F=N-mg which he changed into ma=N-ma (he changed gravity into ma) and got N=2ma=2W. I don't feel like this is right but if someone could tell me if it is or isn't that be great.

15. Dec 10, 2013

### PhanthomJay

it is not right, but you are almost there... you have the net force correct... F_net = 3W/4. Now if you call the support reaction force N, then the two forces acting on the rod are W down (where W = mg) and N up. Since you know that the tangential (linear) acceleration of the rod is down, then W must be greater than N , and thus the net force is W - N. And that is equal to your calculated net force of 3W/4. Now solve for N!

16. Dec 11, 2013

So then the answer would be W/4?

17. Dec 11, 2013

### PhanthomJay

Yes!!(it is usually expressed as mg/4 in SI convention, which is the same thing).

18. Dec 11, 2013