Fahad Hameed

Fahad Hashmi is one of the known Software Engineer and blogger likes to blog about design resources. He is passionate about collecting the awe-inspiring design tools, to help designers.He blogs only for Designers & Photographers.

25 thoughts on “Veritasium “4 Revolutionary Riddles” Video Response

  • September 28, 2017 at 1:46 pm
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    I have the same conclusions but I didn't have one for the train. Although after seeing this video I realized the answer. The shape of the wheel isn't just a circle. It has a bit that goes next to the track so the train doesn't fall off. That bit is lower that the track meaning it will always go backwards when the train is moving forward!

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  • September 28, 2017 at 1:46 pm
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    NERD!! No no i joke, I kissa you face! Decent vid, Good effort, they all stumped me and i was thankful to see the results today so i comment on your reply vid because you only got 100k view holy wow that's.. oops lost in the crowd ;];]
    Thx for heads URLup of his quantum mechanics video, i'll officially never actually understand (apparently, hehe) ;] Stay curious my fellow traveller -Respect

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  • September 28, 2017 at 1:46 pm
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    Dude I think u got the 3rd one wrong, u cant do the average by doing that u, the Vavg = distance/time which makes the problem impossible because he would have to make the 2nd lap in 0 sec

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  • September 28, 2017 at 1:46 pm
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    Don't bother watching, most of the explanations are wrong. Still, I appreciate the effort.

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  • September 28, 2017 at 1:46 pm
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    For the train one i think that the bottom of the wheel would stay the same speed as the ground because as it spins it would be "pushing the ground backwards" but it can only push it back as fast as the wheel itself was rotating. That was my idea for that one though

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  • September 28, 2017 at 1:46 pm
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    Prob 1. Wouldnt a viscous solution only cause it to move slower down the ramp but not cause it to stop?
    For the second problem, there is a counteractive force of the bike moving forward, which is the tension force applied by the string. How does that play into it. (Its not a force perpendicular to the movement of travel, which is usually negligible, but in this case it is opposite to the movement of travel)

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  • September 28, 2017 at 1:46 pm
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    avg speed is the average of speeds? Check your math. V1 = D/T1, Vavg = 2.D / (t1 + t2), Vavg = 2V1 <=> 2.V1 = 2D / (t1 + t2), you need t2 = 0 to satisfy the answer, i.e you have to run infinitely fast. If you completed both laps in the time it took you to run the first, your avg speed would be 2x. But since you only did one lap, you have no time left to reach that avg speed.

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  • September 28, 2017 at 1:46 pm
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    For the fourth riddle wouldn't the components inside the train be the ones moving backwards compared to the motion? I came to this solution because Work can be described as (W=F(force applied)*d(displacement of the object)*cos(theta). Since the objects inside the train only have a force of friction acting upon them, the force felt by the objects will be negative compared to the motion; and since Work can be made on an object that is stationary therefore the only think keeping the objects inside the train from moving back is a force of friction that is parallel to the motion. Since the components inside the train will have a negative force acting upon then, by using (F=m*a) you can say that the objects inside the train have a negative acceleration.

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  • September 28, 2017 at 1:46 pm
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    You got the third problem wrong. The answer is actually V2 = infinite

    I made the same mistake as you at first, problem is that V1+V2 / 2 doesn't take into account that velocity is a relationship between distance and time.
    For instance, if I run the first lap at 5Km/h, according to that formula, I'll need to run the second lap at 15Km/h in order to get an average of 10Km/h, right? —> 5 + 15 / 2 = 10
    Problem is that the two laps are completed at different speeds, which means different distance-time ratios…therefore, that equation doesn't describe what's actually going on, can't divide different ratios by 2 and expect a correct answer.

    So let's say that the circuit is 5Km total, first lap takes 1h (@5Km/h) and the second lap takes 20min (@15Km/h), that means that the total time is actually 1h 20min (or 1,33333333333333333….hours)
    We recalculate the overall speed using the new data:
    D1+D2/T1+T2=V—> (5Km + 5Km / 1h + 0.333333…h = 7.5Km/h)

    Or we could simplify:
    D/T=V —-> (10Km /1.33333333…h = 7.5Km/h)

    As you can see, we're below the desired average of 10Km/h, and it doesn't matter how fast we run the second lap, it will always take time, and that will always force us to divide those 10Km by a number greater than 1.
    The only way to reach the average of 10Km/h is by completing the second lap in no time, which basically requires infinite speed.
    Derek didn't ask for V1+V2/2, he asked for a D2/T2 (or V2) value which would allow for 2V1= D1+D2/T1+T2

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  • September 28, 2017 at 1:46 pm
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    Hey nice tries. Personally I had the same idea for 1, 2 and 3, but I think 4 cant be right, because Veritassium stated the movement should be backwards (!) with respect to the ground (!) and the bottom part of the wheel stands still compared to the ground, right? But probably thats my English understanding or my idea beeing wrong. (would love an answer)

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  • September 28, 2017 at 1:46 pm
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    the rolling cylinder is fluid filled with small holes that allow the CG to adjust slowly
    use energy and all becomes clear

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  • September 28, 2017 at 1:46 pm
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    to solve the bike problem use an energy method
    it is trivial
    if you pulled back on some magic object and it pulled you forward that would in fact be the solution to free energy
    isnt going to happen in a world with our laws of thermodynamics

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  • September 28, 2017 at 1:46 pm
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    draw a section of a train wheel (that is a "section" through the wheel in the plane of the axel showing the rail in section)
    you will see that the wheel has a flange that has a greater diameter than the portion of the wheel bearing on the rail
    you are correct that the diameter of the wheel where it contacts the rail has no velocity
    but the flange in fact moves backwards anywhere within the region below the rail
    draw proper diagrams and you will solve these problems – or go to engineering school
    cheaper and faster to just learn to draw… haha
    have a great day!

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  • September 28, 2017 at 1:46 pm
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    Not sure if anyone has commented on this as I've not read the other comments in full.

    The train riddle. Have you missed the fact a train wheel is not a simple wheel but also has a flange that extends outside the running circumference. This the part that holds the wheel central to track and prevents the wheel from falling off the track.
    As this extends below the stationary point at the rim it will effectively move backwards as it passes the point at which it is directly below the point in contact with the track and the axle. As the axle moves forward and the centre lower rim moves backwards the lower point of the flange will monumentally move backwards.

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  • September 28, 2017 at 1:46 pm
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    I believe your equation for Average Velocity is wrong;

    Average Velocity = Total distance travelled / Total time taken.

    We have 2 equations for it; let's call it V0;

    V0 = 2 V1

    and

    V0 = (2 L ) / (L/V1 + L/V2)

    If you write the two equations out, you will find that both cases only hold true for V1 = V2 = 0.

    Therefore it cannot be done.

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