Ant on a Rubber Rope Paradox

I am sure you know this paradox quite well that's why you are here to read this post but I think you are not able to grasp the solution of the first order differential equation.

Let's forget the calculus solution and do it with intuitive way. It might not provide you the solution but we may have great discussion.

To solve the complex problems like this, break the problem in chunks so start with the shorter version.

An ant is crawling on a stretchable rope at a constant speed of 1 cm/s. The rope is initially 4 cm long and stretches uniformly at a constant rate of 2 cm/s. When will the ant reach to the initial end point (Corner of the 4 cm rope)?

Initially the ant is at the starting point of the 4 cm rope.

After 1 second, the ant travels 1 cm on the rope. Now here consider the rope does not stretch continuously. It means the rope is stretched 2 cm after the end of every second (The stretching time is negligible i.e. a Femtosecond). This thought might change the whole approach but it can be quite easy to understand because if the rope would stretch continuously then it would stretch all the smallest slots of the second as the ant moves ahead. 

So After 1 second, the ant has traveled 1 cm, and the rope is stretched 2 cm uniformly. It means every one cm of the rope is stretched 0.5 cm so complete 4 cm is stretched to 6 cm.

It makes ant's traveled distance 1 cm to 0.5 cm more stretched so after 1 second the ant has traveled 1.5 cm.

After 2 seconds, the ant travels 1 more cm. So the ant is at the 2.5 cm from the start but the rope is stretched again 2 more cm after the end of 2 seconds. This time this increased 2 cm distance will be distributed between existing 6 cm uniformly. It means every one cm of the rope is stretched 0.3333 cm so complete 6 cm is stretched to 8 cm.

 

It makes ant's traveled distance 2.5 cm to '2.5 x 0.33333 = 0.8333'  more stretched so after 2 seconds the ant has traveled 2.5 + 0.83333 = 3.3333 cm and so on

By this way, the ant will reach to the initial end point between 9 and 10 Seconds.

But if you solve this question using calculus, the answer is 12.77 Seconds

L=4 cm (Initial Length of the rope)

v=2 cm/sec (Rope stretching Speed)

u=1 cm/sec (Ant's speed)

 

There is difference in the answers, this is because we took the different approach but it will give you an idea how you will try next time to solve this question for your own satisfaction.

Approach : 2

Initially the ant is at the starting point of the 4 cm rope.

a) After 1 second, the ant travels 1 cm on the rope.

So After 1 second, the ant has traveled 1 cm, and the rope is stretched 2 cm uniformly. It means every one cm of the rope is stretched 0.5 cm. This 1 cm has been traveled by the ant so stretching distance will be (1x0.5)/2=0.25.

Ant will travel 1.25 cm after 1 second.

b) After 2 seconds, the rope is stretched again 2 more cm. This time this increased 2 cm distance will be distributed between existing 6 cm uniformly. It means every one cm of the rope is stretched 0.3333 cm so complete 6 cm is stretched to 8 cm.

If ant would stand still at 1.25 cm then it would have been moved to 1.25x0.3333=0.41666 cm more that is 1.25+0.416666=1.666666667 cm but ant is moving with speed 1 cm/sec so this 1 cm distance traveled by the ant will be stretched with (1x0.3333)/2=0.16666 so the total distance moved by ant after 2 seconds will be 1.666666+1+0.166666 =2.833332 cm

c) After 3 seconds, the rope is stretched 2 more cm. This time this increased 2 cm distance will be distributed between existing 8 cm uniformly. It means every one cm of the rope is stretched 0.25 cm so complete 8 cm is stretched to 10 cm.

If ant would stand still at 2.83333 cm then it would have been moved to 2.83333x0.25=0.7083325 cm more that is 2.8333333+0.7083325=3.5416658 cm but ant is moving with speed 1 cm/sec so this 1 cm distance traveled by the ant will be stretched with (1x0.25)/2=0.125 so the total distance moved by ant after 2 seconds will be 3.5416658+1+0.125 =4.6666658 cm

d) After 4 seconds, the rope is stretched 2 more cm. This time this increased 2 cm distance will be distributed between existing 10 cm uniformly. It means every one cm of the rope is stretched 0.2 cm so complete 10 cm is stretched to 12 cm.

If ant would stand still at 4.6666658 cm then it would have been moved to 4.6666658x0.2=0.93333316 cm more that is 4.6666658+0.93333316=5.59999896 cm but ant is moving with speed 1 cm/sec so this 1 cm distance traveled by the ant will be stretched with (1x0.2)/2=0.1 so the total distance moved by ant after 2 seconds will be 5.59999896 +1+0.1 =6.69999896 cm and so on.

By this way, the ant will reach to the initial end point between 12 and 13 Seconds.

This is an almost same time calculated by calculus solution.

The difference between approach 2 and previous approach is the stretching distance calculation of ant's traveled distance. We took half of the stretching distance in approach 2 and it gets us near the exact answer.

Case II

Now Let's discuss one easy condition in this question. Suppose the ant is at the fixed distance from the initial point. Let's say 1 cm from the initial point but the ant does not move a bit. Only stretching factor of the rope makes it move further.

One thing is clear in this condition, the ant will never reach to the end point.

But it will never be at the 1 cm always so how much will it move.

At 0th second, it is 1 cm from the start.

After 1 second, rope is stretched 2 more cm. This 2 cm is distributed between existing 4 cm uniformly that makes the total length 6 cm. Every cm is stretch to 0.5 cm so the ant will be at 1.5 cm from the start. That is 25% of the total length.
75% remaining.

After  2 seconds, rope is stretch 2 more cm. This 2 cm is distributed between existing 6 cm uniformly that makes the total length 8 cm. Every cm is stretch to 0.3333 cm so the ant will be at 2 cm from the start. That is 25% of the total length. 75% remaining.

After  3 seconds, rope is stretch 2 more cm. This 2 cm is distributed between existing 8 cm uniformly that makes the total length 10 cm. Every cm is stretch to 0.25 cm so the ant will be at 2.5 cm from the start. That is 25% of the total length. 75% remaining.

It means if the ant does not move, The 'distance to cover' ratio does not change and that's why if the ant starts moving very slowly, it will always reach up to end,  if sufficient time is given and stretching speed is constant.

Case III

Now there is another condition. Suppose the rope gets doubled after every second. Speed of the ant is same i.e. 1 cm / second and initial length of the rope is also 4 cm.

What will happen now ? Will it reach up to the end point?

After 1 second, length of the rope is 8 cm. It means increased 4 cm is distributed between existing 4 cm uniformly. Every one cm of the rope is stretched to 1 cm. The ant will be at 2 cm but length of the rope will be 8 cm.

After 2 seconds, length of the rope is 16 cm. It means increased 8 cm is distributed between existing 8 cm uniformly. Every one cm of the rope is stretched to 1 cm. The ant was at 2 cm and now it travels 1 more cm in 2nd second, as the rope is stretched, the ant will be at 6 cm but length of the rope will be 16 cm.

After 3 seconds the ant will be at 14 cm from the start and length of the rope will be 32 cm.

After 4 seconds the ant will be at 30 cm from the start and length of the rope will be 64 cm.

The stretching speed of the rope is not constant, it is accelerating and that's why the ant will never reach up to the end point.

 

 

Mathematics Quiz II - How Many Squares Are In This Picture

Calculate the total number of squares in the following image.

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Answer : 40

Explanation :

If there is an n x n square (number of squares are in horizontal row should be equal to vertical row)

there is a way to calculate total number of squares in it.

Suppose there is a 2 x 2 square

How many squares are in it. If you say 4 then you are not absolutely right. Correct answer is 5. There is also a big square ACIG

Now how do you calculate this answer in mathematical terms.
Total number of squares in an n x n square =  n² + (n-1)² + (n-2)² + ......+ 1²
For 2 x 2 square = 2² + 1² = 4 + 1 =  5

Now suppose there is 3 x 3 square

How many squares are in it. Using above formula, we can easily calculate.

Total number of squares in an n x n square =  n² + (n-1)² + (n-2)² + ......+ 1²
For 3 x 3 square = 3² + 2²  + 1² = 9 + 4 + 1 =  14
So where are these 14 squares.
ABFE, BCGF, CDHG, EFJI, FGJK, GHLK, IJMN, JKON, KLPO, ACKI, BDLJ, EGOM, FHPN and ADPM

Now similar way we can calculate squares in Original question. There is one 4 x 4 square and two 2 x 2 squares in the image.
Total number of squares in a 4 x 4 square = 4² + 3² + 2²  + 1² = 16 + 9 + 4 + 1 =  30
Total number of squares in a 2 x 2 square = 2² + 1² = 4 + 1 =  5
There is another 2 x 2 square in image = 2² + 1² = 4 + 1 =  5
So Total number of squares = 30 + 5 + 5 = 40

Exercise for you : How many squares are in a Chess Board? :)

Interstellar (2014) - Calculations and Analysis

                                             Interstellar (2014)
                                              Welcome to the Fountain
                                           Quench Increase your Thirst

Fact : 1
Miller's Planet to outside observers orbits Gargantua every 1.7 hours. On Miller's Planet, that means the planet orbits ten times a second around Gargantua , which is normally faster than the speed of light. But since the spin from Gargantua caused space to whirl around it similar to wind, Miller's Planet does not travel faster than light relative to its space as the laws of physics say you cannot travel faster than light relative to space, but space itself is not bound by the speed limit. As such, faster than light travel is possible by bending and twisting space. However, Gargantua would have to fill half the sky in order for it to be so close.

Fact : 2
The time dilation on Miller due to the gravitational forces of Gargantua would be tantamount to the planet moving through empty space at roughly 99.99999998% the speed of light. 

Fact : 3
Gargantua’s mass must be at least 100 million times bigger than the Sun’s mass. If Gargantua were less massive than that, it would tear Miller’s planet apart. The circumference of a black hole’s event horizon is proportional to the hole’s mass. For Gargantua’s 100 million solar masses, the horizon circumference works out to be approximately the same as the Earth’s orbit around the Sun: about 1 billion kilometers.

Fact : 4
Miller’s planet is about as near Gargantua as it can get without falling in and if Gargantua is spinning fast enough, then one-hour-in-seven-years time slowing is possible. But Gargantua has to spin awfully fast. There is a maximum spin rate that any black hole can have. If it spins faster than that maximum, its horizon disappears, leaving the singularity inside it wide open for all the universe to see; that is, making it naked—which is probably forbidden by the laws of physics

Fact : 5
Einstein’s laws dictate that, as seen from afar, for example, from Mann’s planet, Miller’s planet travels around Gargantua’s billion-kilometer-circumference orbit once each 1.7 hours. This is roughly half the speed of light! Because of time’s slowing, the Ranger’s crew measure an orbital period sixty thousand times smaller than this: a tenth of a second. Ten trips around Gargantua per second. That’s really fast! Isn’t it far faster than light? No, because of the space whirl induced by Gargantua’s fast spin. Relative to the whirling space at the planet’s location, and using time as measured there, the planet is moving slower than light, and that’s what counts. That’s the sense in which the speed limit is enforced.


QUERIES 
1. How old is Miller’s planet? If, as an extreme hypothesis, it was born in its present orbit when its galaxy was very young (about 12 billion years ago), and Gargantua has had its same ultrafast spin ever since, then the planet’s age is about 12 billion years divided by 60,000 (the slowing of time on the planet): 200,000 years. This is awfully young compared to most geological processes on Earth. Could Miller’s planet be that young and look like it looks? Could the planet develop its oceans and oxygen-rich atmosphere that quickly? If not, how could the planet have formed elsewhere and gotten moved to this orbit, so close to Gargantua?

2. What is the gravitational time dilation equation for Miller's planet? As there is 60000 ratio between time on earth and time on Miller's planet , to balance the equation what should be the distance of planet from Gargantua, angular momentum as it is revolving around very fast spinning object and Mass of Gargantua ?

3. We all know Gravitational Time Dilation does not affect Mann's planet as it is far from Gargantua’s vicinity. But we also know, almost immediately after the Endurance’s explosive accident in orbit around Mann’s planet, the crew find the Endurance being pulled toward Gargantua’s horizon. From this it appears when crew leaves Mann’s planet, the planet must be near Gargantua. Following diagram is the orbit of Mann's Planet.

According to this, what should be the orbital period and orbital velocity of Mann's planet for a person on Earth and a person on Mann's planet.

4. How long Dr. Mann spent time on Mann's planet according to him and according to an observer on Earth(Keep its orbital path in mind)? How long did he spend in hibernation for both observers?

5. Why was Endurance able to receive signal from Earth but Earth was not able to receive signal sent by Endurance.

6. When Cooper left Earth, he was 35 Years old and when he returned, He was 124 years old for Murph. 35 + 2 years for Saturn Journey + 23 years for Miller's Planet Journey + 51 years for Black Hole Journey = 111, Where are 13 Years Missing? How Long was Cooper out for himself?

8. What is the Orbital velocity of Miller's planet for a person on Earth and a person on Miller's planet?

9. What is the age of Brand when Cooper arrives at Edmund's Planet?

10. How long Dr. Laura Miller spent time on Miller's planet according to her and according to an observer on Earth? Lets keep in mind that she died minutes ago before Cooper and Brand reached there.

11. Miller’s planet travels around Gargantua’s billion-kilometer-circumference orbit once each 1.7 hours. Could Rom see it moving very fast from Mothership?

12. If Coop and team would try to communicate with Rom from Miller's planet, how would their communication appear? According to Coop how fast would they get response from Rom and similarly how long would Rom get response from Coop & team?