Find Distance d for Impulse on Stick to Rotate Once

[SothSogi]

Homework Statement

(Sorry for not typing like Latex, I still do not know how)

One has a uniform stick of length l which is held horizontally. At t=0 it is released from rest. At the same time, a sharp upward blow by a force of magnitude F perpendicular to the stick is applied at a distance d for some \delta t. The stick flies then fall under the action of gravity. One has to determine the distance d such that when the stick's center falls back to the same height it started with, the stick rotated exactly once. (Assume F is much larger than gravity when the impulse was delivered) (I guess I can use Delta t in my answer, therefore I do not worry about it)

Homework Equations

torque=r x F

L=I*omega (I is the moment of inertia)

omega=dtheta/dt

The Attempt at a Solution

I got

torque=d*F*Delta t=Delta L

Since at the beginning only F delivered angular impulse, and it was at rest, I have

torque=d*F*Delta t=L_f=I*omega

Now, one revolution implies dtheta=2*pi. So I have

d*F*Delta t=I*[(2*pi)/dt]

But I do not know what to do with that dt.

Any suggestions?

Thanks in advance.

 

[TSny]

There is no picture. Is d the distance that the force is applied as measured from the center of the stick?

Regarding finding the time interval dt, what does this time interval represent? Is there a way to express the vertical velocity of the center of mass of the stick, just after the impulse, in terms of the impulse?

 

[SothSogi]

Yes. d is the distance from the center of the stick to the point where the impulse was applied.

Also, dt comes from dtheta/dt. Since the constrain of 1 rotation makes dtheta=2*pi, I supposed I needed dt.

To find the velocity of the CM after the impulse was delivered I can use F*Delta t=m*v_cm. But what can I calculate with it? The max height it reached before falling under gravity?

Thanks.

 

[TSny]

To find the velocity of the CM after the impulse was delivered I can use F*Delta t=m*v_cm.

Yes

But what can I calculate with it?

Can you relate dt to v_cm,0?

 

[SothSogi]

That would be $v_{cm}=\frac{dx_{cm}}{dt}$. I can substitute this on my "result", then the result does not depend on the impulse, but now I have the $dx_{cm}$.

Thanks for your guide, sir.

 

[TSny]

I'm still wondering if you have a clear interpretation of the meaning of "dt" when you wrote (2*pi)/dt for omega. What particular time interval does dt refer to? Can you describe what happens to the stick during the time dt?

 

[SothSogi]

Since it is the angular velocity, and I wrote $2\pi$, it should be the time it takes to complete exactly one rotation. But you are right, I am still confused about that. (That is why I got stuck with this problem, I still do not understand rotational dynamics very well, although I have solved correctly the other problems of the assignment).

I also have that $v=d*\omega$ for rotational motion, (this time $d$ is the distance) then $F\Delta t=m*d*\omega$. Then I can substitute this into the torque equation and solve for $d$.

 

[TSny]

Since it is the angular velocity, and I wrote $2\pi$, it should be the time it takes to complete exactly one rotation. But you are right, I am still confused about that.

OK, good. So, dt is the time interval that it takes the stick to rotate once. What happens to the cm of the stick during this time?

I also have that $v=d*\omega$ for rotational motion, (this time $d$ is the distance) then $F\Delta t=m*d*\omega$. Then I can substitute this into the torque equation and solve for $d$.

I don't think you will need this.

 

[SothSogi]

Thanks for your help.

The CM goes up (due to the impulse), then down (due to gravity), during that time interval. I guess I have to calculate this time interval using the kinematic equations (I also wanted to include torque due to gravity, but I admit I simply do not know how to do that).

Note: Besides solving the problem, I would like to know how can you spot my mistakes so easily, I mean, I am trying to solve this problem like for 2 day or so. I am amazed about how easily you seem to know the answer. Guess I am not made for this hehe.

 

[TSny]

The CM goes up (due to the impulse), then down (due to gravity), during that time interval. I guess I have to calculate this time interval using the kinematic equations

Yes. Try to express dt in terms of v_cm,0.

(I also wanted to include torque due to gravity, but I admit I simply do not know how to do that).

The force of gravity on the stick can be thought of as acting through the cm of the stick. So, it doesn't produce any torque about the cm.

Note: Besides solving the problem, I would like to know how can you spot my mistakes so easily, I mean, I am trying to solve this problem like for 2 day or so. I am amazed about how easily you seem to know the answer. Guess I am not made for this hehe.

I'm an old timer who's seen these types of problems many times over many years. This problem would likely have been a challenge for me when I was taking my first course in physics. Experience is the key. So work lots of problems!

 

[SothSogi]

Thank you very much for your help. I always get depressed with these problems hehe. I guess practice makes perfect.

So the time it takes to go up and down is $t=\frac{2v_{cm,0}}{g}$. With this, one can write $d*F\Delta t=I\frac{2\pi}{dt}=I\frac{2g\pi }{2v_{cm,0}}$, then, with the linear impulse $v_{cm,0}=\frac{F\Delta t}{m}$, then

$d*F\Delta t=I\frac{gm\pi }{F\Delta t}$

Substituting the moment of inertia for a rod of mass $m$ gives me $d=\frac{1}{12}g\pi\left(\frac{ml}{F\Delta t}\right)^2$

The units are consistent, but I do not know, the result seems weird, does not it?

 

[TSny]

$d=\frac{1}{12}g\pi\left(\frac{ml}{F\Delta t}\right)^2$

This result, as well as your work, looks correct to me.

The units are consistent

Good thinking to check that!

but I do not know, the result seems weird, does not it?

What do you find weird about the result?

 

[SothSogi]

Thank you very much! I noticed the "Homework helper-Gold Member-2017 award" title you have! OMG, I didn't know who was helping me :)

With respect to the units, I learned doing that saves you a lot of time, and some of my classmates do not have that habit, but I consider it simply crucial.

And what I find weird about the result is that, well, one is asked to find a distance, then one ends up with $g$ and $\pi$, and a weird ratio squared, hehe. I mean, that is the beauty of physics, but as I am not an expert I think I expected something like a fraction of $l$ in my result.

Thank you very much, once again!

 

[TSny]

With respect to the units, I learned doing that saves you a lot of time, and some of my classmates do not have that habit, but I consider it simply crucial.

That's a great observation.

And what I find weird about the result is that, well, one is asked to find a distance, then one ends up with $g$ and $\pi$, and a weird ratio squared,hehe. I mean, that is the beauty of physics