# Bug on a hinged square metal frame

1. Jan 11, 2014

### Saitama

1. The problem statement, all variables and given/known data
A square metal frame in the vertical plane is hinged at O at its centre as shown in the figure. A bug moves along the rod PN which is at a distance $l$ from the hinge, such that the whole frame is always stationary, even though the frame is free to rotate in the vertical plane about the hinge. Then, the motion of the bug is simple harmonic with the time period

A)$2\pi\sqrt{l/g}$

B)$2\pi\sqrt{2l/g}$

C)$2\pi\sqrt{4l/g}$

D)$2\pi\sqrt{l/(2g)}$

2. Relevant equations

3. The attempt at a solution
There are two forces acting on the frame, friction (f, along PN) and normal reaction due to bug (N', vertically downwards). Since the frame must not rotate, $fl=N'(\sqrt{2}a-2l-x) \Rightarrow f=N'(\sqrt{2}a/l-2-x/l)$, where $x$ is the displacement of bug from P and $a$ is the side length of square frame.

The bug moves from P to N and the frictional force on bug is directed from N to P. Applying Newton's second law on bug, $m\ddot{x}=-f=N'(x/l-2-\sqrt{2}a/l)$. This doesn't turn out to be SHM.

Any help is appreciated. Thanks!

#### Attached Files:

• ###### hinged square.png
File size:
11.1 KB
Views:
70
2. Jan 11, 2014

### TSny

It might be nicer to let $x$ be measured horizontally from the midpoint of PN.

What force acts on the bug to accelerate the bug? What does this tell you about the magnitude of $f$ acting on the rod? [Edit: Sorry. At second glance, it looks like you have already answered this.]

What can you say about the magnitude of the Normal force N'?

Last edited: Jan 12, 2014
3. Jan 12, 2014

### Saitama

What's wrong with my $x$?

I get these equations according to your $x$:
$$f=\frac{N'x}{l}$$
$$-m\ddot{x}=-f \Rightarrow m\ddot{x}=\frac{N'x}{l}$$
It's mg. :)

4. Jan 12, 2014

### TSny

There's nothing wrong with your $x$, you'll still get SHM. But I think you'll see why choosing the origin in the middle of the rod makes the expressions simpler.

You almost have it. What would need to be changed in your equation for $\ddot{x}$ to get SHM. Can you justify the needed change? (Hint: Does your symbol $f$ stand for the force that the bug exerts on the frame, or the force that the frame exerts on the bug?)

5. Jan 12, 2014

### Saitama

I am not sure but I got the LHS in the following manner:
$$F=m\frac{d^2}{dt^2}\left(\frac{a}{\sqrt{2}}-l-x\right)$$
Let the terms inside the parentheses be $u$. Since $u$ decreases with time, I have to place a minus sign i.e
$$F=-m\ddot{u}$$
Am I correct?

6. Jan 12, 2014

### TSny

If $x$ is the position of the bug relative to the origin at the center of the rod, how do you express the acceleration $a_x$? I don't understand why you wrote $\frac{d^2}{dt^2}\left(\frac{a}{\sqrt{2}}-l-x\right)$.

To get the signs right, draw a picture with the bug at some arbitrary positive value of $x$. Decide which way the friction $f$ must act on the rod. What does that tell you about the direction of the friction force on the bug?

7. Jan 12, 2014

### ehild

It is simpler with the torque equation: The torque of the external force is equal to the time derivative of the angular momentum of the system. The torque is from the weight of the bug; the angular momentum changes as the bug moves.

ehild

8. Jan 12, 2014

### TSny

Yes, that's a very nice way to do it.

9. Jan 13, 2014

### Saitama

The bug moves from P to N. Let $u$ be the distance between point P and bug at time $t$. Then, acceleration of bug is $d^2u/dt^2$. Since we measure $x$ horizontally from hinge, $u$ is related with $x$ by
$$u=\frac{a}{\sqrt{2}}-l-x$$
Differentiating twice wrt time,
$$\ddot{u}=-\ddot{x}$$.
So from Newton's second law,
$$m\ddot{u}=-f \Rightarrow m\ddot{x}=f$$

Please tell me if anything is still unclear.

Hi ehild! :)

This is what I did:

The torque on bug about the hinge is $fl$. Since the friction acts along NP, the frictional torque is clockwise. The magnitude of angular momentum of bug is $L=mvl$ where $v$ is the velocity of bug at any instant. Since $dL/dt$ is anticlockwise, we have
$$-fl=\frac{dL}{dt} \Rightarrow -fl=ml\frac{du}{dt} \Rightarrow m\ddot{x}=f$$
I still end up with the sign problem. :(

10. Jan 13, 2014

### TSny

We must not be defining $x$ in the same way. I am thinking of an x-axis running along the rod PN with origin $x=0$ at the midpoint of the rod. Positive x is toward the right. As $x$ increases, the bug moves farther away from P.

Sorry for not understanding your definition of $x$.

11. Jan 13, 2014

### Saitama

I am still confused with the signs. :(

I am going with your definition of $x$. Let the length of rod be $k$. Then, the displacement of bug from P is given by $s=k-x$ and acceleration of bug is hence $d^2s/dt^2$. Am I right so far?

#### Attached Files:

• ###### bug on frame.png
File size:
12.6 KB
Views:
54
12. Jan 13, 2014

### TSny

If you are going with my definition of x, surely s must increase as x increases. Remember, I choose positive x direction to be toward the right. Your picture seems to be taking x positive to the left.

13. Jan 13, 2014

### haruspex

The algebra will be simpler if you consider displacement from the centre of the rod. If x is displacement to the right, the torque from normal force is mgx clockwise. The horizontal acceleration of the bug is $\ddot x$, so the force on the rod from the bug is $m\ddot x$ to the left.

14. Jan 14, 2014

### Saitama

Ah, I think I get it now, I was being really foolish.

The displacement from P is $s=k/2+x \Rightarrow d^2s/dt^2=d^2x/dt^2$. Hence,
$$F=m\ddot{s}=m\ddot{x}=-f$$
This turns out to be SHM, thanks a lot TSny!

Continuing with ehild's approach, I had
$$-fl=\frac{dL}{dt} \Rightarrow -fl=\frac{d(mvl)}{dt}=ml\frac{dv}{dt}$$
Since
$$\frac{dv}{dt}=\ddot{s}=\ddot{x}$$
Hence, I get the same equation. Thanks a lot ehild for this nice method.

But I can't get the answer using the $x$ I originally took in my attempt, I am ending up with the same error as in OP.

15. Jan 14, 2014

### ehild

I meant the torque equation for the whole system bug+frame. Friction is an inner force between bug and frame, it does not count.

I choose a system of coordinates with horizontal x axis. pointing to the right and vertical y axis, pointing upward.

The only external force is gravity. Its torque with respect to the hinge is τ=-xmg The angular momentum of the whole system is that of the bug as the frame does not move: L=lmv

$\tau=-xmg$, $L=Mlv$, $\tau= \dot L \rightarrow -xmg=l m \dot v \rightarrow -xmg=lm\ddot x$ which corresponds to the equation of SHM.

When writing up torque and angular momentum with respect to the centre of the frame do not forget that they are cross products. L=rxmv, τ=rxF.

ehild

#### Attached Files:

• ###### bugonframe.JPG
File size:
7.8 KB
Views:
48
Last edited: Jan 14, 2014
16. Jan 14, 2014

### Saitama

Hi ehild! :)

Yes, I understand the torque approach but I am not able to get the answer using the $x$ I originally took, can you please help me point out the error?

In my original approach, I set origin at P and positive x-axis towards right (or PN).

17. Jan 14, 2014

### haruspex

So going back to the OP:
$\sqrt{2}a-2l$ is the distance PN, so $\sqrt{2}a-2l-x$ is the distance of the bug from N.
Don't you want the moment of N' about O? That would be $N'(a/\sqrt{2}-l-x)$, anticlockwise, yes? (When $x > a/\sqrt{2}-l$ the sign changes to make it clockwise.)
If f is the horizontal force on the frame, the equation $fl = N'(a/\sqrt{2}-l-x)$ implies f is measured positive to the left.
No, according to the above usage, f on the bug must be positive from P to N.

18. Jan 15, 2014

### ehild

With the coordinate system shown in the picture, the hinge is at (b/2, l) (b is the length of the rod). So r=(x-b/2)i-l j. τ=rX(-mg j)=-mg(x-b/2)k, and L=rXmvi =lmvk. τ=dL/dt--> -(x-b/2)mg=lmd2x/dt2.

Going back to forces acting on the frame and bug separately: The bug exerts normal force on the rod, it is equal to mg (downward) and its torque is mg(b/2-x), anti-clockwise (b is the length of the rod).
The bug exerts some frictional force f on the rod, and the torque of that force compensates the torque of the gravitational force. f l+ mg(b/2-x)=0
The bug exerts f force on the rod, the rod exerts -f force on the bug. So the motion of the bug corresponds to ma=-f=mg/l (b/2-x)
If the bug starts from P (x=0) and its initial speed is 0, it will move according to x=bg/(2l)(1-coswt)

ehild

#### Attached Files:

• ###### bugframe.JPG
File size:
8 KB
Views:
49
Last edited: Jan 15, 2014
19. Jan 16, 2014

### Saitama

Hi ehild!

Sorry for the delay in reply.

Thank you very much for the explanation, I think I see the error I had with the signs. The direction of forces was alright in my working (I think) but I messed up with the torques, thank you again. :)