# Can adding weight to something make it tilt slower?

Let's say we have 2 sticks, both a meter long. We put both of them on the ground vertically. To the top of one of them we attach a weight. Then we tip both of them over and let them fall.

Assume that there is no air resistance and that (the lower ends of) the sticks do not slide against the floor.

Which one falls first?

Would I be correct in guessing that the one with the weight falls slower due to having a higher center of gravity?

EDIT: In case you're wondering where this question came from, I was pondering whether having a passenger on the back of a motorcycle would increase or decrease the time it takes to tip over in the case balance is lost (at very low speed).

• Can adding weight to something make it fall slower? Well, you + a parachute weighs more than you alone... May 8, 2015 at 18:11
• That this is true is close to an explanation for the falling chimney effect (i.e. you need to exert force to slow it down): myweb.lmu.edu/gvarieschi/chimney/chimney.html May 8, 2015 at 18:25
• @MasonWheeler Did you hear about the physicist who jumped out of a plane, opened his parachute, and ended up splattering on the ground in a bloody mess? He omitted the effect of air resistance... May 9, 2015 at 10:15
• Excellent question, but the title itself is a little misleading. Falling implies linear acceleration when the question (and the accuracy of the answers) depend instead on angular acceleration. Also, the angular acceleration is only half the story; the overall instability of the system i.e. the likelihood of a fall initiating is actually increased with a higher center of gravity. (According to my intuition and rudimentary understanding, at least. Apologies if I'm mistaken.) May 9, 2015 at 19:04

I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like:

The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity.

To write down the equation of motion for the rod we use the rotational analogue to Newton's second law:

$$T = I\frac{d^2\theta}{dt^2}$$

where $T$ is the torque and $I$ is the moment of inertia, and rearranging this gives:

$$\frac{d^2\theta}{dt^2} =\frac{T}{I} \tag{1}$$

If the length of the rod is $\ell$, then the torque is:

\begin{align} T &= mg\frac{\ell}{2}\cos\theta + Mg\ell\cos\theta \\ &= (\frac{m}{2} + M)g\ell\cos\theta \end{align}

$$I_{rod} = \frac{m\ell^2}{3}$$

and assuming our weight can be approximated as a point mass its moment of inertia is:

$$I_{weight} = M\ell^2$$

and the total moment of inertia is just the sum of these two:

$$I = \frac{m\ell^2}{3} + M\ell^2$$

And to get our equation of motion we just substitute our expressions for $T$ and $I$ into equation (1) to get:

$$\frac{d^2\theta}{dt^2} = -\frac{(\frac{m}{2} + M)g\ell\cos\theta}{\frac{m\ell^2}{3} + M\ell^2}$$

There's a minus sign because as I've drawn the diagram the angle $\theta$ decreases with time so the angular acceleration is negative. With some rearranging we get:

$$\frac{d^2\theta}{dt^2} = -\left(\frac{m + 2M}{m + 3M}\right) \frac{3g\cos\theta}{2\ell} \tag{2}$$

Now actually solving this equation of motion would be hard, but we don't need to solve it to answer your question. The left hand side of equation (2) is the angular acceleration, and if we increase the magnitude of the angular acceleration the rod falls faster while if we decrease it the rod falls slower. So your question simplifies to:

If we increase the mass M does the magnitude of the angular acceleration increase or decrease?

On the right hand side everything outside the brackets is independent of $M$, so we just need to answer whether the term in the brackets increases or decreases if we change $M$.

This is easy to answer because we have a term $2M$ on the top of the fraction and a term $3M$ on the bottom, and obviously $3M \gt 2M$. So if we increase $M$ the fraction in the brackets decreases, and therefore the magnitude of the angular acceleration decreases.

So the answer is that attaching a mass to the top of the rod does indeed make it fall more slowly.

• Could you take that shortcut of reasoning about the influence of $M$ earlier with $I$? Increasing $M$ increases $I$ and $I$ is the "resistance to motion" (roughly speaking), thus increasing $M$ increases said resistance.
– Name
May 8, 2015 at 17:54
• @Name: I strongly advise against the taking of shortcuts unless you're absolutely confident you understand the underlying physics. For example in this case increasing $M$ does indeed increase $I$, but it also increases $T$. Is it obvious which increase is greater? I'm not sure I would have wanted to bet on it before I'd done the maths. May 8, 2015 at 18:07
• "obviously $3M>2M$. So if we increase $M$ the fraction in the brackets decreases" This argument is faulty. $\frac{m+2M}{m+3M}$ decreases as $M$ increases, while $\frac{m+2M}{2m+3M}$ increases. Your conclusion is correct, but the explanation fell apart at this step. May 8, 2015 at 18:18
• @JohnRennie: Your argument shows that the fraction converges to a value less than one. The step showing whether this is an increase or a decrease is simply missing. May 8, 2015 at 18:40
• The caveat is that at some point the contact is going to start slipping, and later it might separate altogether. This changes the rotational speed profile as those transitions occur at different angles for each situation. May 8, 2015 at 22:25

I assume that you're imagining two sticks whose bases stay stuck still to the ground as they tip over. In this case, what we should do to calculate the tipping rate (I wouldn't exactly say "falling"-- I think it's misleading) is consider the torque applied to each stick as it tips.

What the answer really comes down to is the struggle between torque (the rate of change of angular speed) and moment of inertia (a measure of resistance to torque) as mass is added in different places on the sticks.

The moment of inertia $I$ is equal to $m r^2$ for a point mass, while torque scales linearly with the force applied perpendicular to the radial direction (basically in a rotating direction, not pushing into the center of the stick). In this case, that torque force comes from gravity. In addition, torque scales linearly with the distance from the centre of rotation at which the force is applied. $$\tau = F \cdot r \cdot \sin \theta$$ ...where $\theta$ is the angle between the direction in which force is applied, and the direction outwards from the center of rotation to that same point.

In this scenario, torque will be larger for the stick with the mass on the end because both the center of mass of the stick and its total mass will be larger. However, the moment of inertia (which resists torque) will also be larger. In fact, due in part to $I$ scaling with $r^2$, in this case it actually trumps the increase in torque and results in slower falling for the stick with the mass on the end. I will now prove this mathematically:

I'll assume that each stick is an infinitely thin rod of uniform mass-per-length $m$. The mass of each stick is then $m$. Let the mass attached to the top of stick 2 be $M$. Stick 1 has no mass attached.

The torque applied to each stick as a function of angle from vertical is: $$\tau = h\cdot xg \cdot \sin \theta$$ ...where $h$ is the height of the center of mass of each stick when stood upright, and $x$ is the total mass of the stick. For the two sticks, this gives us: $$\tau_1 = 0.5 \cdot mg \cdot \sin \theta_1$$ $$\tau_2 = \frac{M + 0.5m}{M + m} \cdot (M + m)g \cdot \sin \theta_2$$

Simplfying:

$$\tau_1 = 0.5m \cdot g \cdot \sin \theta_1$$ $$\tau_2 = (M + 0.5m)g \cdot \sin \theta_2$$

In order to find the rate of change of angle, we divide torque by the moment of inertia. Without going into detail of how they're calculated, I'll just give the moments of inertia:

$$I_1 = m/3$$ $$I_2 = m/3 + M$$

Then we can calculate the angular acceleration of each stick as $\tau / I$:

$$\ddot \theta_1 = \frac{0.5m}{m/3} \cdot g \cdot \sin \theta_1$$ $$\ddot \theta_2 = \frac{(M + 0.5m)}{m/3 + M} g \cdot \sin \theta_2$$

Simplifying:

$$\ddot \theta_1 = 1.5 \cdot g \cdot \sin \theta_1$$ $$\ddot \theta_2 = \frac{(M + 0.5m)}{m/3 + M} g \cdot \sin \theta_2$$

The stick that tips over faster is the one with the coefficient in front of $\sin \theta$ of greater magnitude. So we'll look at the ratio of the two expressions for the same angle $\theta_1$ and $\theta_2$. What we're doing is comparing the angular acceleration of each stick at any particular angle $\theta$: $$\frac{\ddot \theta_1}{\ddot \theta_2} = \frac{1.5 \cdot (m/3 + M)}{M + 0.5m} = \frac{0.5m + 1.5M}{0.5m + M} = 1 + \frac{0.5M}{M + 0.5m} > 1$$

We can see then that $\ddot \theta_1 > \ddot \theta_2$ and so we conclude that the first stick without the mass $M$ attached on the end will be the one that tips over most quickly.

I haven't strictly shown that stick 1 will hit the ground first, but it should be obvious from the fact that it accelerates faster at every angle during its tip-over.

I just want to formulate John Rennie's answer in a slightly different and more general way.

For any object of angular momentum $I$, the angular acceleration $\dot \omega$ is given by

$$\dot\omega = \frac{\Gamma}{I}$$

where $\Gamma$ is the torque.

Now in the case of an object with masses $m_i$ distributed along the length at distance $\ell_i$ from the axis of rotation,,

$$I = \sum m_i \ell_i^2$$

and the torque, at an angle $\theta$ to the vertical, is given by

$$\Gamma = \sum m_i \ell_i g \sin\theta$$

For both of these, the sum over discrete masses could easily be the integral over a continuous mass distribution. It doesn't change what follows.

The question then becomes - what is the minimum distance $d$ at which I have to add a mass $m$ so that $\dot\omega$ decreases? Using my expression for moment of inertia and torque, this happens when

$$\frac{\sum m_i \ell_i g \sin\theta}{\sum m_i \ell_i^2}\gt \frac{M d g \sin\theta + \sum m_i \ell_i g \sin\theta}{M d^2 + \sum m_i \ell_i^2}$$

Rearranging:

$$\left(Md^2 + \sum m_i \ell_1^2\right) \cdot \sum m_i \ell_i\gt \left(Md + \sum m_i \ell_i \right) \cdot \sum m_i \ell_i^2\\ Md^2 \sum m_i \ell_i \gt Md \sum m_i \ell_i^2\\ d \gt \frac{\sum m_i \ell_i^2}{\sum m_i \ell_i}$$

This tells us exactly where we have to put a mass in order to make the object fall more slowly. If the initial object is a point mass at the end of a stick of length $\ell$, then $d\gt \ell$. If the initial object is a rigid rod of length $\ell$, then the numerator is the moment of inertia $I = \frac13 m \ell^2$, and the denominator will be $\frac12 m \ell$, so we need

$$d \gt \frac{\frac13 m \ell^2}{\frac12 m \ell} = \frac23 \ell$$

It follows that putting a mass at any point above $\frac23$ of the way up the stick, including the top, will make it fall more slowly, and adding a passenger to a motorbike may, from this simple perspective, increase the stability.

Having ridden motorbikes, I can tell you that it depends a lot on the passenger... My wife for one does not behave like an ideal point mass (and if I called her that I'd be likely to get punched).

Yes you are right. There is a resist in gravity because of an applied weight. Therefore the acceleration of the beam falls slightly slower depending on the mass of the object. Unless nothing are holding them together, both beams would fall at the same rate except the applied weight on the top would fall slower. For an example, If we are in an elevator and the elevator drops downward, We wouldn't move. But that's where gravity comes in. Newtons 2nd law states that objects in motion stay in motion. Its as almost as we are concealed in a room without gravity. Until the elevator (like the beam) comes to a complete stop. Then by multiplying our mass with earth's gravity we accelerate quickly to the ground. (like the applied weight on the beam) F(n)=mF(g) and F(n)=ma These are simple and helpful equations.

• PSA: If you disagree with this answer, downvote it, don't flag it. May 8, 2015 at 18:21
• can someone explain to me why this answer is not good? May 9, 2015 at 17:09
• @Ooker Well I didn't downvote, but looking at it mostly it just seems confused/confusing. Lots of ideas jumbled together and not clearly explained. May 9, 2015 at 21:30
• @Ooker It is mostly incoherent, and seems to imply that in a vacuum, the feather hits the ground before the hammer. May 10, 2015 at 17:25