A rod so long!

Classical Mechanics Level 5

Consider a thin rod having length same as earth's radius $R=6371 Km$ and uniformly distributed mass $m = 20 Kg$ . Its lower end is just above earth's surface as shown in the figure. The rod is hinged about $O$

Find the time period of small angular oscillations of the rod in seconds .

Details and assumptions

Value of $g$ near surface of earth = $9.8 m/s^2$ .
You may neglect friction and air resistance.

The answer is 4136.42.

1 solution

Jatin Yadav
Feb 19, 2014

Since it is quite difficult to draw neat and complicated force diagrams at MS Paint, showing all distances forces, angles etc., hence, rather than analysing torque I would be analysing energy here, even though I used force diagram.

Let $M$ be mass of earth. Clearly, potential energy = $U = \displaystyle \int_{0}^{R} \frac{-GM\frac{m}{R} dx}{r}$ ,

Using cosine formula, $r = \sqrt{x^2+4R^2 - 4Rx \cos \theta}$

Hence, $U = \frac{-GMm}{R} \displaystyle \int_{0}^{R} \frac{dx}{\sqrt{x^2+4R^2 - 4Rx \cos \theta}}$

Torque $\tau = \frac{-dU}{d \theta} = -2GMm \displaystyle \int_{0}^{R}\frac{xdx \sin \theta}{(x^2+4R^2 - 4Rx \cos \theta)^{3/2}} \approx -2GMm \int_{0}^{R} \frac{x dx}{(2R-x)^3} \theta$

= $-\frac {GMm}{2R} \theta$

Now, $\tau = \frac{mR^2}{3} \ddot \theta$ .

Hence, $\ddot \theta = - \frac{GM}{R^2} \frac{3}{2R} \theta = -\frac{3g}{2R} \theta$

Hence, time period $\boxed{T = 2 \pi \sqrt{\frac{2R}{3g}} \approx 4136.42 s}$

Note: The integral $\displaystyle \int_{0}^{R} \frac{x dx}{(2R-x)^3}$ can easily be calculated by substituting $2R-x=z$

I worked with forces. But I am not getting the correct answer. Can you tell what's wrong?

The force on a mass element at a height $h$ from the surface of the Earth is:

$dF = \dfrac{GM\frac m R dh} {(h+R)^2}$ .

So, the total force on the body is $\int dF$ = $\dfrac{mg}{2}$ by taking the limits as $h = 0$ and $h = R$ .

This force acts on the center of mass (not the center of gravity) which is at a distance $\dfrac R2$ from the hinge. Taking the torque about that point, we get:

$\dfrac{mg}{2} \dfrac R2 \sin \theta \approx \dfrac{mg}{2} \dfrac R2 \theta = \dfrac{mR^2}{3} \alpha$ . That gives the time period as :

$T = 2\pi\sqrt{\dfrac{4R}{3g}}$

What's wrong?

Parth Thakkar - 7 years, 3 months ago

Taking the force at centre of mass is valid only if the gravitational field is uniform. Here it varies as $1/r^2$ .

Pranav Arora - 7 years, 3 months ago

The force $mg$ acts on the center of gravity. On the centre of mass, only $mg/2$ is acting.

The center of mass is given by:

$h_{cm} = \dfrac{ \int h dm } {\int dm }$ So, $h_{cm} \int dm = \int h dm$

Differentiating twice wrt time, we get: $a_{cm} \int dm = \int a dm$ That is: $M_{T}a_{cm} = \int dF = F_{net}$ .

Clearly this holds for all forces. That's what I have done here!

Parth Thakkar - 7 years, 3 months ago

@Parth Thakkar – The equation which you have given above regarding linear acceleration of center of mass is really not relevant here. In order to really use that equation you would have to take into account the force at the point of suspension as well (since it is an external force).

However, if you are only considering torque, then the force at the point of suspension would not contribute to the torque about this point.

The total weight is mg/2 and it acts at the center of gravity (not center of mass). You could translate that force to the center of mass, however there would also be an additional moment = (mg/2)*d, where d is the distance between the center of gravity and the center of mass.

Reaber John - 7 years, 3 months ago

@Reaber John – A correction: The additional moment is (mg/2)*d sin theta.

Reaber John - 7 years, 3 months ago

@Reaber John – Hmm! Now THAT is a huge error - not considering the force at the hinge. Thanks for pointing it out. Thanks a lot!!

For center of mass-center of gravity business:

If I somehow consider the force at hinge too (don't ask how), then that force (the total: vectorial sum of gravity and hinge-force) would act at the center of mass, right?
And please tell if the following is correct:

If we consider the gravitational force alone, then it will act at the center of gravity (which happens to be different from CM here) and not the center of mass.

Parth Thakkar - 7 years, 3 months ago

@Parth Thakkar – Regarding the first question, the answer is no.

First observe that in a reference frame translating with the center of mass, the body has the same angular acceleration $\displaystyle{ \ddot{\theta} }$ . Further, it can be shown that the following equation holds : $\displaystyle{\tau_{c.m.} = I_{c.m.} \ddot{\theta} }$ where $\displaystyle{\tau_{c.m.}}$ is the net torque of all external forces about the center of mass and $\displaystyle{ I_{c.m.}}$ is the moment of inertia about the center of mass.

Thus, if the net force (including force at the hinge) were to pass through the center or mass, then the torque $\displaystyle{\tau_{c.m.}}$ would be zero and thus $\displaystyle {\ddot{\theta} }$ would be zero from the above equation. But since $\displaystyle{ \ddot{\theta} }$ is not zero (except when $\theta = 0$ ), so the net force does not pass through the center of mass (except when $\theta = 0$ ).

By the way, since we have the differential equation for $\theta$ , it can be solved and $\theta$ can be known as a function of time provided we are given the initial conditions. Then we could find the net acceleration (both centripetal as well as tangential acceleration) of the center of mass. Then using the equation(s) for the motion of center of mass, we could even find the force(s) acting at the hinge (since all the other forces i.e. gravitational forces are known ). This hinge force will be a function of $\theta, \dot {\theta}$ and $\ddot{\theta}$ .

Regarding second question, the answer is yes. Perhaps one should use the expression instantaneous center of gravity( since it might depend on $\theta$ when $\theta$ is not small ).

Reaber John - 7 years, 3 months ago

@Reaber John – With respect to your explanation of my first doubt: Whatever you said makes sense. So, is it that the equation 'I' had derived - $F_{net} = M_{total}a_{cm}$ doesn't mean that the net force acts at the center of mass? Because that's what the impression I've had through out.

Oh and btw, a bigger question: when we say that this force acts at that point, what we mean is that the torque of that force about that point is zero. Right?

Parth Thakkar - 7 years, 3 months ago

@Parth Thakkar – Yes, the derivation doesn't imply that the net force passes through the center of mass. Yes to the second question as well.

Reaber John - 7 years, 3 months ago

@Reaber John – Thanks for answering my questions with such patience and such great details!

Parth Thakkar - 7 years, 3 months ago

The center of gravity is at a distance of 2R/3 from point O. ( I got this by considering the displaced position of the rod and equating moments due to only forces perpendicular to the rod about the point O to the moment due to the perpendicular component of total weight of the rod about point O.)

Taking torque about point O, we get

$-\displaystyle {\frac{mg}{2} \frac{2R}{3} \frac{3 \theta}{2} }$ = $\displaystyle {\frac{mR^{2}}{3} \ddot{\theta} }$ .

That gives the required time period as

$T = 2 \pi \displaystyle{ \sqrt {\frac{2R}{3g}}}$ . The expression $\displaystyle{ \frac{3 \theta}{2} }$ in the above is the sum of angles (since angles are small the sine of angle is replaced by angle itself) from the point O and from the center of the earth to the center of gravity of the rod.

I made two assumptions (1) that the weight i.e. $\displaystyle{ \frac {mg}{2} }$ is independent of $\theta$ for small angles and (2) that the total weight acts toward the center of the earth from the center of gravity. They need to be checked.

Reaber John - 7 years, 3 months ago

I don't understand how you got the center of gravity. You say that you got it by equating: 1. (perpendicular wrt rod) forces to the rod about the point O 2. (perpendicular wrt rod) component of total weight of rod about point O.

The second force is $mg/2\sin\theta$ which acts at the center of gravity (that lies at an unknown distance, which has to be found, from O).

What is/are the first force(s)?

Parth Thakkar - 7 years, 3 months ago

@Parth Thakkar – We are equating two moments about point O.

The first moment is due to a distributed gravitational force acting at each point. Since this force is known at each point, we can calculate the total moment about the point O. It is clear that only forces perpendicular to the rod will contribute to the moment.

We intend to replace the distributed gravitational force by a single force equal to the vector sum of the distributed gravitational forces and acting at the center of gravity (which is to be found). This replacement force acts at such a point (i.e. center of gravity) that it gives the same moment about point O, as that due to the distributed gravitational forces. Again it is clear that only the perpendicular ( to the rod) component of the replacement force contributes to the moment about O.

Thus, equating the the two moments the unknown location of center of gravity can be obtained.

We could have chosen any other points besides O to evaluate the moments and hence find the center of gravity. But this point O is convenient since only forces perpendicular to the rod contribute to moment about O.

BTW, the second force is $\displaystyle{\frac{mg}{2} \sin( \theta + \varphi_{c.m.})}$ where $\varphi_{c.m.}$ is the angle OCP where P is the center of gravity. See the figure by Anish Puthuraya.

The first moment is the same as obtained by Anish and is :

$\displaystyle{ T_{1} = 2 GMm \theta \frac{1}{4R} = \frac {mgR}{2} \theta}$

The second moment can obtained as

$\displaystyle { T_{2} = \frac {mg}{2} x_{c.m.} \sin( \theta + \varphi_{c.m.} ) }$

Equating the two moments, and using

$\displaystyle {\sin ( \theta + \varphi _{c.m.}) = \theta + \varphi_{c.m.} = \frac {2R}{2R - x_{c.m.}} \theta}$

we get the location of center of gravity as $x _{c.m.}= 2R/3$ .

Finding the center of gravity is not an advantage in solving the given problem since we would have to first obtain the net moment $T_{1}$ anyway. However, the purpose of this was to show that we could consider the net force acting at center of gravity and obtain the same result for the time period.

Reaber John - 7 years, 3 months ago

@Reaber John – By "net force acting at center of gravity" I mean the net gravitational force.

Reaber John - 7 years, 3 months ago

@Reaber John – I used subscripts c.m. by mistake ! it should be c.g. for center of gravity.

Reaber John - 7 years, 3 months ago

Thanks. Very nice solution...
Btw I prefer Adobe Photoshop to MS Paint.

Anish Puthuraya - 7 years, 3 months ago

Hey can you tell what's wrong with what I've done? This one has been bugging me since many days!

Parth Thakkar - 7 years, 3 months ago

I think you approximated too much (you have to take into account $\theta$ as well to get the approximations precise)...So, maybe you got the force wrong.

alt text

I shall take the total torque instead of the total force...

From the diagram,
$x\theta = r\varphi$

Taking the torque about the hinge,
$dT = \frac{GM\frac{m}{R}dx}{r^2}\times\sin(\theta+\varphi)\times x$

Also, using Cosine Rule,
$r^2 = 4R^2+x^2-4Rx\cos\theta$
Thus,
$\displaystyle r^2 \approx (2R-x)^2$
$\displaystyle r \approx (2R-x)$

Using all the above equations, (substitute $\displaystyle\varphi$ in terms of $\displaystyle\theta$ )
$dT = \frac{GMm}{R}\frac{r+x}{r}\theta\frac{xdx}{(2R-x)^2}$

Substituting the value of $\displaystyle r$ ,
$dT = 2GMm\theta\frac{xdx}{(2R-x)^3}$

Integrating from $0$ to $R$ , we get,
$\tau = 2GMm\theta\times\frac{1}{4R}$

We know,
$\tau = \frac{mR^2}{3}\alpha$

Thus,
$\alpha = \frac{3GM}{2R^2}\theta = \frac{3g}{2R}\theta$

Hence,
Time Period $\displaystyle = 2\pi\sqrt{\frac{2R}{3g}}$

Anish Puthuraya - 7 years, 3 months ago

@Anish Puthuraya – Ohk! So it was the problem of approximating too much. Thanks for that! Probably even in calculating force, I need to consider $\theta$ . Hmm...finally, after so many days :D

Parth Thakkar - 7 years, 3 months ago

@Parth Thakkar – Dont worry, even I did not solve the problem...After 3 tries, I found this solution, but it was obviously too late.

Anish Puthuraya - 7 years, 3 months ago

@Anish Puthuraya – Cool...so a guy like you too gets problems wrong, eh?! I know, too much flattering...

Oh and btw, a friend of mine used the formula: $T = 2\pi\sqrt{\dfrac{I}{mgd}}$ where $I$ is the Moment of inertia about the hinge, and $d$ is the distance between the CM and the point of hinging. That gave the answer straight away! Funny thing is, that 'formula' is derived assuming $g$ to be a constant. Just saying!

Parth Thakkar - 7 years, 3 months ago

@Parth Thakkar – Yeah, its just a coincidence...Pretty interesting though..

Anish Puthuraya - 7 years, 3 months ago

@Anish Puthuraya – Indeed, this was the solution I had, using forces.

jatin yadav - 7 years, 3 months ago

@Anish Puthuraya – Yes this is the way i got,,, damn i was constantly getting ln(2) in my equations as i took the angle to be too small and then calculating the torque,, fortunately i remembered the question relating to the rotating rod in magnetic field of Jatin yadav and realised that the problem must lie in over approximation,,

Mvs Saketh - 6 years, 10 months ago

woah,, that was a really nice one,,, easier than considering torques

Mvs Saketh - 6 years, 10 months ago

Same way !!!!

A Former Brilliant Member - 4 years, 2 months ago

A rod so long!

The answer is 4136.42.

1 solution

0 pending reports