# Which block reaches the floor first?

There are two blocks, each starting at the top of an incline. The particular inclines are depicted in the image below.

The height through which the blocks fall is the same, the table lengths are the same, and the height of the tables is the same. The question asks which block hits the floor first. One solution argues that the first block hits the ground first. This is because the the slope of that incline is steeper, so the block reaches the bottom of the incline faster.

I remain unconvinced by this argument. I am not sure how one can make an argument without doing a more detailed analysis on the inclines. The information is not provided to perform this analysis, so I assume it is not truly necessary.

The reason I am unconvinced by this answer is that, despite the slope of the first incline being steeper at the beginning, I feel this means the slope must be less steep near the end of the trajectory. I believe this because the entire vertical and horizontal distance traversed by the block is the same in both cases. If the slope is steeper at the beginning for the first block but steeper at the end for the second block, then how can we conclude anything?

• I agree with you that a deeper analysis will make it much clearer. But one way to think about the argument as stated is, that when the steep decline happens in the beginning and there never is any upwards incline later again, then the object will traverse the rest of the track at a higher speed than on the second case. Thus making it to the end quicker. If the steep decline happens later on the track as in the second case, then the speed will be lower for a longer while, so the total duration might be longer. Commented Apr 4 at 8:34
• A related problem is one of shortest time with a trajectory along the brachistochrone curve. Commented Apr 4 at 9:34
• What do sprinters do in the $100$ metre dash? Start off slowly and then speed up mid-race? Commented Apr 4 at 9:42
• Friction effect is included in the task or surface is assumed to be frictionless ? Also if friction is non-negligible,- what's a total sliding curves length ? Because in case friction is important,- we can construct such curves (stair-case like) where we would artificially slow-down block in each "pit-hill" to make it slide with constant average speed, given that we can build long enough track. Seems like task is under-specified. Commented Apr 4 at 10:36
• Block 1 looks like it will drop if let go. Block 2 looks like it needs a force applied. Commented Apr 4 at 12:52

Among all the sloping surfaces the inverted Cycloid as a Brachistochrone has minimum time to slide down, proved in Variational mechanics. The first slope line differs only slightly from the inverted Cycloid, so is faster than the second convex sloped incline which is way off from the Cycloid, so takes more time.

Brachistochrone

• This curve $~ y(x)=\frac {A}{x+0.01}~$ is not Cycloid and it look like curve 1
– Eli
Commented Apr 6 at 19:47

Looking at the pictures I would also claim that the first one is faster. My argument would be, that both curves seem to reach a similar point at about $$\frac{2}{3}$$ of the horizontal and about $$\frac{7}{8}$$ of the vertical distance to their starting point. Looking at the route prior to reaching that point we see that the left curve is significantly below the right curve at all points. That means that on the first $$\frac{2}{3}$$ of the horizontal the Block rolling down the left curve will be a lot faster then the block on the right curve, while the traveling distance up to that point seems similar. At that point both objects will already have reached around $$\frac{7}{8}$$ of their maximum kinetic energy, which means they reached even more then $$\frac{7}{8}$$ of their maximum speed. Therefore, what ever differences in steepens of the routes may be present to the right of that point will make a very small contribution to the overall travel times of the blocks compared to the differences picked up on the route prior to reaching that point.

Generally acceleration at the start of the curve will cause a larger contribution to the reduction of the travel time, then the same acceleration at a later part of the curve, because the block will travel a larger distance at the increased speed.

To solving this problem it is not necessary knowing the solution of the Brachistochrone problem. It would be like to kill a fly with a gun!

It is possible to answer by observing that the profile of the first incline has grater derivative than the second one at each value of the ascissa $$x$$.

The form of the profiles is such that the map $$y=y(x)$$ is invertible so that we can describe the profile either as $$y=y(x)$$ or $$x=x(y)$$ and $$x'(y) = 1/y'(x)$$. This permits to perform some explicit computation.

By energy conservation (I assume that the slopes are frictionless and $$U(y):= mgy$$) $$U(y)+ \frac{m}{2} {\bf v}(y)^2 = U_0 = U(d+h)$$ That is $$\frac{m}{2} \left(\left(\frac{dy}{dt}\right)^2 + \left(\frac{dx}{dt}\right)^2\right)= U_0- U(y)\:.$$ Namely $$\frac{m}{2} \left(\left(\frac{dy}{dt}\right)^2 + \left(\frac{dx}{dy}\frac{dy}{dt}\right)^2\right)= U_0-U(y)\:.$$ Equivalently, $$dt = \frac{\sqrt{\frac{m}{2}} \sqrt{1+ \left( \left(\frac{dy}{dx}\right)^{-1}\right)^2}}{\sqrt{U_0-U(y)}} dy$$ As a consequence, the considered lapse of time amounts to $$T = \int_{d}^{h+d} \frac{\sqrt{\frac{m}{2}} \sqrt{1+ \left( \left(\frac{dy}{dx}\right)^{-1}\right)^2}}{\sqrt{U_0-U(y)}} dy \:.$$ The only difference in the integrands in the two cases is the derivave $$\frac{dy}{dx}$$. From the figure, it seems clear that $$\left|\frac{dy_1}{dx}\right| \geq \left|\frac{dy_2}{dx}\right|$$ and thus $$\frac{\sqrt{\frac{m}{2}} \sqrt{1+ \left( \left(\frac{dy_1}{dx}\right)^{-1}\right)^2}}{\sqrt{U_0-U(y)}} \leq \frac{\sqrt{\frac{m}{2}} \sqrt{1+ \left( \left(\frac{dy_2}{dx}\right)^{-1}\right)^2}}{\sqrt{U_0-U(y)}}$$ When integrating, we have $$T_1 \leq T_2$$

Here's another way to think about it (the other answers seem to use very complicated concepts on a relatively simple problem to me). Draw the velocity vs. time graphs for both situations. You should end up with something like this, and it should be intuitively obvious:

Where the top figure is for situation 1 and the bottom figure is for situation 2.

The distance travelled is the area under the curve, and block 1 obviously travels a larger distance than block 2, for the same amount of time. Therefore, unless you have reason to believe that the slope in situation 1 is longer than the slope in situation 2, block 1 should reach the bottom first.

The question asks which block hits the floor first. One solution argues that the first block hits the ground first. This is because the the slope of that incline is steeper, so the block reaches the bottom of the incline faster.

This problem can be simplified by comparing two ramps of different slopes.

At the bottom of both ramps, the object has the same speed. The object with the steeper slope, however, reaches that speed in a shorter time. It's important to note that the second ramp occupies a smaller area then ramp one in the rectangle formed by $$xh$$.

From basic kinematics, the time needed to traverse the horizontal distance $$x$$ for the first ramp is:

$$t = \sqrt{\frac{s_1}{2g sin \theta}}$$

The time to travel the same horizontal distance $$x$$ for the second ramp has two parts: The time to traverse the second ramp:

$$t_1 = \sqrt{\frac{s_2}{2g sin \phi}}$$

and the time to traverse the remaining horizontal distance b:

$$t_2 = \frac{b}{v_f}$$ where $$v_f$$ is the velocity at the end of the ramp. If $$t' = t_1 + t_2$$, then the ratio of the two times for each ramp is: $$\frac{ t'} t = \frac{(\sqrt{\frac{2 s_2}{gsin\phi}}+\frac{b}{v_f})}{\sqrt{\frac{2 s_1 }{gsin\theta } } }$$

This can be simplified since $$s_1 = \frac{h}{sin\theta}$$ and $$s_2 = \frac{h}{sin\phi}$$ and $$v_f = \sqrt{2s_1gsin\theta}$$

This leaves:

$$\frac{t'}{t } = \frac{ sin\theta}{sin\phi} +\frac{bsin\theta}{2h}$$ with $$b = x-hcot\phi$$

Since $$tan\theta = \frac{h}{x}$$ then,

$$\frac{t'}{t} = \frac{sin\theta}{sin\phi} + \frac{sin\theta(cot\theta - cot\phi)}{2}$$

The ratio of times are now expressed only as a function of the two angles or ramp slopes.

While it's not obvious, if $$\phi > \theta$$, then t > t' which supports the argument that the path with largest average slope will be the faster path. Also, the longer path may be the faster path. The path length of second ramp is longer than the path of the first ramp.

The graph has $$\theta$$ fixed at 10° while $$\phi$$ starts at 10° and increases. There are two competing effects wrt. ramp two:

1. The increase in acceleration due to increasing slope.
2. The increase in path length due to increasing slope.

These effects are minimized when $$\phi - \theta = 45°$$

Note that the ramp that occupies the smallest area over the distance x and a height of h will cover the distance x in the least time.

Looking at your diagram, it is clear the curved ramp for block one occupies less area then the second multi-curved ramp. So block 1 arrives first.

This short video illustrates the concept.

https://youtu.be/iogVVja1MYY?si=m1NB5NyYvHP4tvMO

The object in situation 1 wins the race. This is because it spends are greater proportion of the trajectory with a large velocity.

As an extreme example, imagine the second situation is one where the trajectory is horizontal from the start to just before the end of the table. To state this clearly, if the table has length $$\ell$$, imagine the situation where the second trajectory is a straight horizontal line up to $$\ell-\epsilon$$ for very small $$\epsilon$$ and then plunges rapidly from $$\ell-\epsilon$$ to $$\ell$$. In this case, the horizontal part of the trajectory would take basically infinite time since the objects are not given any initial velocities, and the horizontal motion will not provide any additional kinetic energy dues to loss in height.

Actually the optimal shape of the curve - obtained from variational calculus - is a cycloid, and it looks very much like situation 1. You'd have to do the calculation to convince you of that though.

• This curve $~ y(x)=\frac {A}{x+0.01}~$ is not Cycloid and it look like curve 1
– Eli
Commented Apr 6 at 19:50
• @Eli sorry but I don't get your point. The optimal trajectory is known to be a cycloid. Of course I realize the curve of situation 1 is not necessarily a cycloid, but one is only asked to compare situations 1 and 2. The tradeoff between how steep you have to descend vs how long the path is expressed mathematically but clearly situation 1 is closer to the cycloid than situation 2. Commented Apr 6 at 21:08
• .Sure the optimal, but who said it is the optimal trajectory?
– Eli
Commented Apr 7 at 7:21
• @Eli sorry I simply do not follow your comment. You can solve the brachistochrone problem and you will get a cycloid. In your own (now deleted) solution set $ds=dy\sqrt{1+(dx/dy)^2}$ and then $v=\sqrt{2gy}$ and solve the resulting Euler-Lagrange equation. The solution is $x=-\sqrt{\frac{y}{c}-y^2}+\frac{1}{c}\arccos(\sqrt{cy})+c_2$ for $c,c_2$ integration constants, which is a cycloid. Commented Apr 8 at 1:33

The final trajectory of the blocks as they leave the table appears to be horizontal, so the height of the table off the floor is immaterial. Both blocks will take the same time to fall from the table to the floor. If the second slope was an inversion of the first, with a near vertical trajectory at the edge of the table, things might be very different.
So, the only question is which block reaches the table first. As other answers have pointed out, the brachistochrone curve is the fastest possible way to descend, and the first curve is very similar to such a slope and will definitely be the faster.
If there is no friction, both blocks will have the same velocity when they leave the table. The blocks have to travel a vertical distance and a horizontal distance. Block 1 will be traveling this horizontal distance at high speed. Block 2 will travel a large portion of this horizontal distance at low speed. Hence, block 1 will take les time.