As the reaction to my previous physics puzzle was somewhat mixed, I'll try again. I found this in an online physics book.

You hold a broom as shown in the following picture:

broom puzzle

The head of the broom is resting on the ground, and you are holding the handle end a couple of feet off the ground. You are also holding a bunch of keys at the same height as the tip of the broom handle.

You let go of everything at the same time. What happens?
A) The keys and the broom handle hit the ground at the same time.
B) The keys hit the ground first.
C) The broom handle hits the ground first.

Explain your answer.

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    $\begingroup$ @OmegaKrypton I guess so. I added the Real tag. $\endgroup$ – Jaap Scherphuis Oct 22 '19 at 10:42
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    $\begingroup$ Imho this should be migrated to the physics site for a definite answer. $\endgroup$ – Guntram Blohm Oct 22 '19 at 16:25
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    $\begingroup$ As the starting angle of the broom has not been stated, there would appear to be more than one correct answer, or, the answers given are incomplete. $\endgroup$ – Weather Vane Oct 22 '19 at 16:33
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    $\begingroup$ You accepted a wrong answer. In fact, the problem is ill-defined, and depends on the mass distribution and angle. $\endgroup$ – Jeffrey Oct 22 '19 at 17:31
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    $\begingroup$ @WeatherVane In the question I say the end is two feet off the ground, and with a normal sized broom of about 5 feet you get an angle just below 25 degrees, very similar to what the picture in the question depicts. Broom handles almost always have a uniform mass distribution, and just like nearly every physics problem it is perfectly reasonable to assume uniform density without having to explicitly state so in the question. The mass of the head of the broom has virtually no effect since it is on the ground already. $\endgroup$ – Jaap Scherphuis Oct 22 '19 at 18:22

My instinct is that:

C. The broom handle hits first

This is based on:

The net effect of gravity acts through the centre of mass. As the centre of mass of the broom handle is half way along the handle, this is the part of the broom handle that will accelerate at 1g, i.e. 9.81m/s2.
This means that the tip of the broom handle will accelerate at twice the middle of the broom, i.e. at 2g. Whereas the keys will just accelerate at 1g. Therefore the tip of the broom will hit the ground in half the time that the keys will.

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    $\begingroup$ You have good instincts, as this is the correct answer. The difference in speed is quite noticeable, and I highly recommend people try it out for real and see for themselves. A small amount of energy is taken up by the rotation of the broom, and if there is friction by the sideways movement of the broom, but at the kind of angle shown in the question those are small enough to make very little difference. $\endgroup$ – Jaap Scherphuis Oct 22 '19 at 13:54
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    $\begingroup$ So do you agree that there must be some starting angle when they both hit the ground at the same time? $\endgroup$ – Weather Vane Oct 22 '19 at 16:16
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    $\begingroup$ Maybe is it the answer? dev.physicslab.org/… $\endgroup$ – Conifers Oct 22 '19 at 16:44
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    $\begingroup$ This is incorrect. It depends on the broom angle. It is only true for broom close to the ground (low angle). If the broom was up at a angle of 89 degrees, visibly, the key would hit first. $\endgroup$ – Jeffrey Oct 22 '19 at 17:28
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    $\begingroup$ @Conifers page 2 of your linked example says the critical angle is 35.3° and although it says minimum angle I think it should say maximum angle as the broom plainly lands last when nearly vertical. $\endgroup$ – Weather Vane Oct 22 '19 at 17:40

If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.

This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify the properties of the broom, but I think these are a reasonable interpretation for a physical model.

Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.

Height Velocity Acceleration


I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters). Their values don't really matter, they just change the constant in $ \ddot{\theta} = - c \cos \theta$, which only affects the time scale.

This formula, also derived on this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.

The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.


I doubt that such a second-order differential equation with $cos$ has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.

I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.

  • $\begingroup$ Pro is here!, And this answer should be accepted! (I still can't figure out somewhere, physic is too hard to me :( ) $\endgroup$ – Conifers Oct 23 '19 at 4:56
  • $\begingroup$ This puzzle seems easier than the site's(but both are hard for me :(, I can't even write an equation...) due to need consider more about the ball's normal force pressured on the rod, and this puzzle key & broom fall separately. $\endgroup$ – Conifers Oct 23 '19 at 5:15
  • $\begingroup$ Thanks for the proper answer xnor :). Angular acceleration is beyond my knowledge set. $\endgroup$ – AndyT Oct 23 '19 at 13:35

Reductio ad absurdum

Suppose the broom is very nearly upright (vertical) and the keys are level with its top.
It can be seen that the broom handle has to move sideways.
Even if the acceleration of the broom tip (sideways) were the same as the keys, it has a longer path to travel - nearly $\pi / 2 = 1.57$ times the distance travelled by the keys.
But to make it worse, the broom's acceleration is initially very small.

Now suppose the broom is very nearly flat (horizontal) and the keys are again level with its top.
This time, they will take almost the same time to fall, as both their acceleration, and distance to travel, are almost the same.
But the broom will still take longer. Because of the slight angle, it has to move a teensy bit sideways, and still travels in an arc and not in a straight line.

The only arrangement when they take the same time, will be when they are both already touching the floor! At all other times, the keys will arrive first.

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    $\begingroup$ Nice logic. Your reductio ad absurdum is very convincing. I'm not sure that it holds for lower angles though. $\endgroup$ – AndyT Oct 22 '19 at 10:20
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    $\begingroup$ Gravity is the only force pulling on the key; the broom gets pulled down by the same gravity, but also experiences a negating force where it touches the floor. So the sum of the forces on the broom is a smaller pull down which causes it to fall slower, meaning it will always reach the floor later. $\endgroup$ – Guntram Blohm Oct 22 '19 at 16:28
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    $\begingroup$ You can create another absurd case: the mass fully in the center of the broom. 0 inertia moment. The center of mass has half the distance to travel, the broom rotates as needed, without slowing down the broom. Then, the broom arrives first. Conclusion: the problem is ill-defined and depends on the inertia moment, length and angle $\endgroup$ – Jeffrey Oct 22 '19 at 17:33
  • $\begingroup$ @GuntramBlohm - But the upward force on the broom head will tend to rotate the broom handle tip downwards... $\endgroup$ – AndyT Oct 23 '19 at 13:34
  • $\begingroup$ "their acceleration ... are almost the same" is not true. You can't assume the tip of the broom has acceleration "g" since it's part of a rigid body. The net acceleration of the entire broom is less than g, but its center of mass starts lower. $\endgroup$ – aschepler Oct 24 '19 at 13:41

Assuming the end of the broom doesn't slide along the floor and neglecting air friction, I believe that:

A is correct for an initial angle of the broom at 60 degrees. B is correct for initial angles greater than 60 degrees. C is correct for initial angles less than 60 degrees.

My reasoning was:

When the broom was almost upright it would start to fall very slowly allowing the keys to hit the ground first. When the broom was almost horizontal the end of the broom would hit the ground first because the center of the handle is the center of gravity and will fall at 9.81 (m/s^2). So the end broom will fall twice as fast.

A more technical explanation:

I equated the kinematic equations for the distance to fall and equated the time until impact. Combined the equations and solved for the angle of the broom. The exact angle calculation is too complex for Puzzling, but the question was good and thought provoking.


So long as the starting angle is close to that shown, the broom end hits the ground first. The question isn't ill-defined so much as requiring a bit of understanding about real-world broom handle lengths and how being a couple of feet off the ground to start with will make that initial angle much less than 47 degrees.

Reason: The easiest way to think of it is that the centre of mass of the broom (roughly its middle) drops at "close enough to"* the same rate as the keys, but the middle is starting off at half the height of the keys, so the broom has an advantage. The end that started at 2ft off the ground will fall at roughly twice the rate of the middle.

*But against that advantage is the component of the force on the head of the broom (that touches the ground) that opposes the force of gravity to some extent - if the broom was perfectly vertical it would entirely cancel the gravitational force... if the broom were horizontal the force would only be rotating the broom; for angles encountered in this question the acceleration of the middle of broom is close enough to that of the keys for the double-rate advantage to let it win the race.


For the moment, consider just the broom handle and the level at which it becomes horizontal: 1 - the tip of the handle and it's midpoint reach there at the same moment: the tip has travelled twice as far but at twice the speed. 2 - the keys must reach that level first.
The only force on the keys is it's weight and it's acceleration is g. On the broom, gravity pulls down, but the ground pushes up on the lower end. So the net force on the handle is less than it's weight and the acceleration of it's CoG must be less than g. The keys beat the CoG and hence the tip. 3 - At that point the keys are travelling faster than the handle CoG and are ahead of it.

For the extra distance required by the thickness of the head, it may be possible for the handle tip to catch up, as it is travelling aout twice as fast as the CoG when it passes horizontal. But if the handle started well above the horizontal, it's CoG is travelling much slower than the keys at this horizontal position, because it's acceleration has always been much less than g. I haven't done the analysis, but I suspect it requires a large broomhead and starting the handle nearly horizontal, if it is to be possible at all.


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