Why Does the Rod Rotate to 67 Degrees After the Clay Ball Impact?

[kanav70]

Homework Statement

A 75 g, 30-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 10 g ball of clay traveling horizontally at 2.5 m/s hits and sticks to the very bottom tip of the rod. To what maximum angle, measured from vertical, does the rod (with the attached ball of clay) rotate?
I know the answer should be 67 degrees but I'm not getting that.
m=mass of clay M=Mass of rod
L=length of rod v=initial velocity of clay ball

Homework Equations

Conservation of Angular momentum: mv(L/2)=Iw
Conservation of Energy: 1/2 Iw^2=mgh

The Attempt at a Solution

Using center of mass I can determine that the center of mass of the rod+clay system is:
[(0.15)(0.075)+(0.01)(0.3)]/0.085=0.1676

The moment of inertia should equal the moment of inertia of the rod plus the moment of inertia of the ball of clay:
ML^2/12 + m(L/2)^2=0.0007875

Now I can solver for w:
mv(L/2)=Iw
4.76=w

Now I can use conservation of energy:
1/2 Iw^2=(M+m)gh
1/2 Iw^2=(M+m)g(0.1676-0.1676coso)
20.6=o

 

[Simon Bridge]

Nice. Do you have a question?

 

[kanav70]

Nice. Do you have a question?

Well my question is why am I not getting the correct answer? Is there a mistake in how I approached the problem?

 

[haruspex]

I get the same for the angular speed just after impact.
There's not enough detail in your working from there to spot your error.
Anyway, there is no need to find the common mass centre, there's an easier way to relate the PE gain to the angle.

 

[kanav70]

I get the same for the angular speed just after impact.
There's not enough detail in your working from there to spot your error.
Anyway, there is no need to find the common mass centre, there's an easier way to relate the PE gain to the angle.

I get the same for the angular speed just after impact.
There's not enough detail in your working from there to spot your error.
Anyway, there is no need to find the common mass centre, there's an easier way to relate the PE gain to the angle.

If you are getting the same angular speed then I probably messed up the conservation of energy.
1/2 Iw^2=(M+m)gh
1/2 (0.0007875)(4.76)^2=(0.085)(9.8)(0.1676-0.1676coso)
0.008921=(0.085)(9.8)(0.1676-0.1676coso)
0.008921/(0.085)(9.8)=(0.1676-0.1676coso)
0.01071=(0.1676-0.1676coso)
-(0.01071-0.1676)/0.1676=coso
-(0.01071-0.1676)/0.1676=coso
0.936=coso
cos-1(0.936)=o
20.6=o

 

[haruspex]

If you are getting the same angular speed then I probably messed up the conservation of energy.
1/2 Iw^2=(M+m)gh
1/2 (0.0007875)(4.76)^2=(0.085)(9.8)(0.1676-0.1676coso)
0.008921=(0.085)(9.8)(0.1676-0.1676coso)
0.008921/(0.085)(9.8)=(0.1676-0.1676coso)
0.01071=(0.1676-0.1676coso)
-(0.01071-0.1676)/0.1676=coso
-(0.01071-0.1676)/0.1676=coso
0.936=coso
cos-1(0.936)=o
20.6=o

No, I think it's your centre of mass calculation. You want it relative to the axle, no? It doesn't look like that is what you calculated.

 

[kanav70]

Thank you I got it!

 

[haruspex]

Thank you I got it!

Good.
But do you see the easier way? The rod's movement does not affect the GPE.

 

[Simon Bridge]

I think it is worth doing a quick and dirty tutorial here.
I'll try not to undermine Haruspex's hints too much: you should verify things for yourself.
... the easier approach involves understanding three main things;

1. Understanding kinetic energy:
kinetic energy is usually given by $K=\frac{1}{2}I\omega^2$ ... but you can use the fact that $L=I\omega$ to get a more useful form for these conservation of momentum problems, vis: $K=L^2/2I$ ... this is the rotational equivalent to the linear $K=p^2/2m$, and it combines the conservation of momentum step with the first half of the conservation of energy step.

That is useful in this case because it allows you to skip the step where you calculated the angular velocity: you know the initial angular momentum and that it is conserved.

There are other cases where alternate forms of the kinetic energy equation are useful - you don't need to memorize them, just know the possibility is there.
The next bit is probably more important...

2. Understanding potential energy:
The gravitational potential energy is the work done to lift the (centre of) mass through a height... it does not matter if all the mass is lifted in one go, or separately: it's the same work. This means that, for the purposes of conservation of energy, we can treat the parts of the rod+ball system separately:
$\Delta U_{tot} = \Delta U_{rod} + \Delta U_{ball}$
This removes the need to calculate the centre of mass, which gave you so much trouble.
Now look at the rod all by itself: when it rotates, how far does the centre of mass move?
So what does that mean about $\Delta U_{rod}$?

This is actually the place where you would have saved the most time, ferinstance: in an exam.
It's how examiners test and reward your understanding of physics: you are penalized if you just go blindly step-by-step through the equations because that takes longer and you have less time to complete other problems.

3. Math discipline:
It is best practise to do all the algebra before plugging numbers into the equations. eg.
If we say the rod has mass M and length r, and the ball has mass m and initial velocity v ... then:
$L=I_{ball}\omega_{ball} = m(r/2)^2(v/(r/2))=\frac{1}{2}mvr$
$I = I_{ball}+I_{rod}=\frac{1}{4}mr^2 + \frac{1}{12}Mr^2$
$\Delta U = mgh = \frac{1}{2}mgr(1-\cos\theta)$ (by geometry)
... and $\Delta K = \Delta U$ (conservation of energy)
So: $$ \frac{(\frac{1}{2}mvr)^2}{2(\frac{1}{4}mr^2 + \frac{1}{12}Mr^2)} = \frac{1}{2}mgr(1-\cos\theta)$$ ... simplify and solve for $\theta$.

Hopefully you can see how the above symbolic form of the equation is easier to troubleshoot?
Notice also how I commented the lines of equations - this reminds me where i got them from and means others can easily figure out what I did wrong.
(Caveat: there may well be a mistake or two in the equations: do not take my word for it - check it yourself.)

The general principle is to stand back and look at what is going on in terms of the physics instead of just looking at the equations, then choosing the form of the equations to suit the problem. The trick, which you will learn by practise, is to treat the mathematics as a language. Those aren't abstract equations, they are descriptions.

Now that is a lot in one go - take it slow.

 

[ehild]

Homework Statement

2. Homework Equations [/B]
Conservation of Angular momentum: mv(L/2)=Iw
Conservation of Energy: 1/2 Iw^2=mgh

The Attempt at a Solution

Using center of mass I can determine that the center of mass of the rod+clay system is:
[(0.15)(0.075)+(0.01)(0.3)]/0.085=0.1676

That is the centre of mass measured from the top end of the rod.

The moment of inertia should equal the moment of inertia of the rod plus the moment of inertia of the ball of clay:
ML^2/12 + m(L/2)^2=0.0007875

You determined the moment of inertia with respect to the axle, at the center of the rod.

Now I can solver for w:
mv(L/2)=Iw
4.76=w

You calculated angular momentum with respect to the axle.

Now I can use conservation of energy:
1/2 Iw^2=(M+m)gh
1/2 Iw^2=(M+m)g(0.1676-0.1676coso)
20.6=o

Using moment of inertia you get the rotational energy with respect to the centre of the rod, which is in rest. The potential energy of the rod does not change when the rod turns around the axis. You need only the position of the clay to take into account which is mgh= mg(L/2)(1-cos(θ))