Angular and Tangential Variables

[7starmantis]

[SOLVED] Angular and Tangential Variables

Homework Statement

A thin rod (length = 1.5m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top if the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: consider using the principle of conservation of mechanical energy.) (b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

Homework Equations

Possibly...
w=vt
mgh = 1/2 mv2
v=st ?

The Attempt at a Solution

I'm really having trouble even setting this one up. I drew a picture of the setup, I know the bar is initially at rest. I'm completely lost on how to even go about it. I've tried the PE=KE equation and canceled out the mass variable which gives me v = 5.4, then I plugged that velocity into the w=vt using a time I got from using 1/2gt2 but that shouldn't really work as its not a free fall problem I think.
Any ideas on how to setup the problem? I think I can figure it out if I can just wrap my brain around the proper setup.

Thanks in advance!

 

[Shooting Star]

Your energy eqn is correct.

Torque = I*Angular accn.

Take torque of the forces acting on the rod about the fixed pt.

 

[7starmantis]

Ok. But how would I find torque without the mass of the rod?
T = LF sin theta...right?
T = (1.5m)(mg) sin (theta) ?

 

[Shooting Star]

How did you get the ang velo? Same way.
(BTW, I presume you meant the ang speed when you said v=5.4.)

 

[7starmantis]

I found angular velocity using w=vt. However its not correct. The t (time) I got from using [tex]\frac{1}{2}[/tex]gt[tex]^{2}[/tex] but that isn't correct as its not a free fall problem, right? So my angular velocity of (5.4)(1.5) = 8.1 rad/s is incorrect. The book gives an answer of 3.61 rad/s.

Am I thinking completely wrong on this one?

 

[Shooting Star]

Let's start again.

The rot KE just before it strikes the floor is the initial grav PE of the rod. If I is the MI of the rod about the fixed pt, and w is the ang speed, then can you write down any eqn?

 

[7starmantis]

Lets see...
mgh=1/2 mv^2
So the KE = [tex]\frac{1}{2}[/tex] I[tex]\omega^{2}[/tex]

Does the PE = Igh ?

so Igh = [tex]\frac{1}{2}[/tex] I[tex]\omega^{2}[/tex]

Wouldn't I cancel leaving gh = [tex]\frac{1}{2}[/tex][tex]\omega^{2}[/tex]

Then I could just solve for [tex]\omega[/tex] ?

 

[Shooting Star]

The grav PE of a body wrt a plane is mgh, where h is the height of the CG of the body above the plane. This would be equal to Iw^2/2. Can you find w now?

 

[7starmantis]

I'm not sure I understand why Iw^2 is divided by 2 in your last post.

For a rod setup this way, I = 1/2 ML^2 ?
However, I'm missing the mass. So I don't think I could find w just yet, right?
I'm sorry, I'm really not trying to be purposefully dull.

 

[Shooting Star]

The expression for rotational KE about a point is Iw^2/2, when the MI of the body about that point is I. It's a formula, like mv^2/2 for translational KE.

For a rod, the MI about one of its ends is mL^2/3, where L is the length of the rod.

Now can you find w, as I said in my last post?

 

[7starmantis]

Ok, my teacher used the equation 1/2 Iw^2 but that's the same as Iw^2/2. Sorry about that. So, mgh = Iw^2/2 ?

Then, I = mL^2/3...
So we get the equation as...

mgh = [tex]\frac{(mL^2/3)w^2}{2}[/tex]

I can if I had the mass of the rod to insert for m in the equation. However the problem said to ignore the mass...
So the full equation should be...

gh = [tex]\frac{(L^2/3)w^2}{2}[/tex]

If so, then yes, I can find w.
w = 6.26

However the answers in the book have the angular velocity as 3.61 rad/s
and the angular acceleration as 6.53 rad/s^2

 

[Shooting Star]

What is the height h of the CG of a rod?

 

[7starmantis]

Hmm...didn't think of that.
It would be W[tex]_{1}[/tex]X[tex]_{1}[/tex] / W[tex]_{1}[/tex] ?

But the rod is oriented vertically so there is no distance and weight is not given, right?

 

[Shooting Star]

The CG of a uniform rod is always at its mid point. So, in the above formula, put L/2 in place of h, and then find w again.

 

[7starmantis]

Ah...man I feel stupid. Thank you so much for your help, I think I really understand what was going on now.

Lets see...

g(L/2) = [tex]\frac{(L^2/3)\omega^2}{2}[/tex]

so that gives [tex]\omega[/tex]=4.43 which is still different from the book answer of 3.61. However sometimes the book gives an answer a bit different from the right one. This is a bit larger than the normal difference though. Does that look right to you?

 

[Shooting Star]

I read your first post again. The rod is massless. There's a mass at the top end of the rod. Do the calc again. It's much easier.

 

[7starmantis]

Heh, ok so just remove the mass component (I) from the equation?
I've done it as gh=w^2/2 that's still not right.

i'm sorry I'm a bit confused I think.

 

[Shooting Star]

Grav PE = mgL. Rot KE just before hitting the ground = (1/2)Iw^2. I=mL^2. So,

mgL = (1/2)mL^2*w^2 => w= 3.61.

 

[7starmantis]

wow, I made this so much harder than it was supposed to be.

I really appreciate your time and help.

Thanks alot!

 

[Shooting Star]

I'm sorry that I made a mistake while reading the problem. On the brighter side, you would be able to do the other kind of problem from now on. Find the angular accn now.

 

[7starmantis]

No problem, yeah that was free info for me!

Ok, so...

[tex]\alpha[/tex] = 3.61 / .55s

[tex]\alpha[/tex] = 6.56

So does that mean that the equation I used for finding the time is correct? If so, how? doesn't the rigid stature of the rod change the time it takes to fall through the distance?

 

[Shooting Star]

No, this is incorrect.

For finding ang accn, look at what I'd said before -- torque etc.

 

[7starmantis]

Ok, so Torque = I * Angular Acceleration

We don't know Torque or I because I involves mass. We can use the same rationale as before and say I = mL^2 but we still are missing mass.

Couldn't we use w[tex]^{2}[/tex] = w[tex]_{o}[/tex][tex]^{2}[/tex] + 2 [tex]\alpha[/tex] ([tex]\theta[/tex])

And use 1/4 2[tex]\pi[/tex] as [tex]\theta[/tex] ?

 

[Shooting Star]

That formula you've given is valid for uniform ang accn.

The torque of the weight of the mass at the point before striking the floor is force*dist from pt of rotation = mg*L. I remains the same as above, i.e., mL^2.

Can you complete the eqn now and find the value with proper units?

 

[7starmantis]

Well since the rod is massless we cannot find mg

the problem did say to ignore the mass so do you simply remove m from the equation?
Giving just gL or (9.8)(1.5m) = 14.7

So 14.7 = 2.25 a
a = 6.53 rad/s^2

Hah, ok so why can we just drop mass out of an equation and still be ok?

 

[Shooting Star]

The problem did not say to ignore the mass at the top of the rod, only the mass of the rod.

Did you not notice that in our earlier calculations, the mass just canceled out from both sides?

All right, now that you have got the correct answers, I recommend a thorough revision of rotational motion.

 

[7starmantis]

Ah, yes I see it now.

Thank you so much, I think I'll go back and read that section and rework those problems again.

Thanks again!
P.S. I have one other problem on here if you want to take a look (its half solved)
https://www.physicsforums.com/showthread.php?t=200199