Can you clear up these 4 rotation-related riddles?

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I got here throughout these 4 physics puzzles through the years in discussions with Neil deGrasse Tyson (riddle 4: which half(s) of a shifting prepare are going backwards with respect to the bottom?), Simon Pampena (riddle 2: run round a monitor twice, the primary time slowly, the second time a lot sooner in order that the common for the 2 laps is twice the pace of the primary lap). Someone tweeted me a video of the thriller cylinder rolling down the ramp in riddle 1 (sorry I am undecided who it was). Riddle three a couple of bicycle going ahead or backward when it is backside peddle is pulled again was dropped at me by a variety of folks and I respect all of their assist!

Filmed by Raquel Nuno.

Thanks to everybody on the Palais de la Decouverte! I’ve had this footage for 5 years and am solely lastly releasing it now. I wished to speak about the best way grass grows on a spinning turntable however I could not find the footage…

source

1. It contains a spinning gyroscope. How it works – Gyroscopes work on principles of magic.

2. The cycle will go backwards. The torques generated by the pull is equal to the one experienced by the back wheel, but larger radius means smaller force at the rim of the wheel, which limits the maximum forward friction to be less than that produced by the pull. Net backward force.

3. Not possible (infinite speed). Twice the average speed means covering twice the distance of the lap in the same time, but that time is already up, leaving zero time for the next lap.

4. Still thinking…

Number 3 is completely possible. V2 has to be 3 times V1. Why? The formula would be ((V1+V2)/2)=2*V1 lets say we go 10 km/h the first lap (yes I know, fast, but it's just an example) then V1 is 10 and 2 times V1 is 20, lets replace that in the equation and say that V2 is our X, the ecuation would be ((10+X)/2)=20. Now solve for X.

1) I think fluid like oil or honey in the container

2) The bike should forward if it doesn't fall over first. I just imagine me sitting on the back peddling from rest. I would apply the similar force as the string I think.

3) (v1+v2)/2=2v1. That makes v2 = 3v1 so the second lap must be 3 times as fast as the first.

4) Trains have wheels where part of the wheel hugs the inside of the rail. This part sits below the point that the wheel is resting on. The part below this section is moving backwards I think. I think I'll have to draw it out and edit this post if I'm wrong.

For number 4 we can try to be clever and say something like "the smoke always goes backwards" but if you want a calculated answer, then, specific parts of the wheels always go backwards. This is a question about cycloids, In a cycloid curve, a point inside a radius always moves backwards during a time interval. Can we find such points on a train? Yes we do: on the rims of the trains wheels. The rims have a larger radius than the circle rolling on the track.

1. Partially filled with some sort of fluid, probably one that is more viscous than water

2. Forward at first as you pull the peddle back as the gears will provide greater forward momentum, then when the pedal is parallel with the ground, the string will begin to decelerate the bike as the peddle is no longer turning to provide acceleration to counter the pull on the bike. Then it will fall over since it is going so slow.

3. This is impossible. If it takes one hour to go one lap, and you then decide you wish to go two laps in one hour, you can't because you already spent your one hour on the first.

4.The bottom half of the wheels are always moving backward with respect to the train. Though I think they are going at the least only 0 with respect with respect to the ground. However, since the video was about rotation, I am still going to say that there is some part of the wheel that is going back words with respect to the ground. Basically I am assuming that some part of the wheel has a larger radius than the part that is touching the track. If so, that part would have a negative velocity with respect to the ground at the bottom of its arc.

1 honey or a similar viscous liquid

2 it'll stay put

3 twice as fast as the first lap

4 the exhaust

1.) sand?

2.) assuming your body remains stationary and your arm is the only force pulling the string, the bicycle will move forward until the pedal has reached 9 o'clock. After 90 degrees of rotation the pedal will increase the distance from your body and the bike. Additionally the pedal would have to be pushed pass 90 degrees to move the bike forward

3.) Velocity for lap 2 would have to be 3 times greater than lap one

4) the top of the wheel will move in the direction of movement. Additionally, so will the "wheel lever"

1) Filled with oil. This generates viscous friction slowing down the cylinder.

2) It might depend on the gear your in. High gear, goes backward. Low gear, goes forward. The mechanical arm from the gear will yield to a reaction either greater or lower than the applied forces resulting in an unbalanced total force (displacement).

3) Second lap at 3*V1. (V1+V2)/2=2*V1 –> V1+V2=4*V1

~~–~~> V2=4*V1-V1=3*V14) The upper part of the wheel goes backward with respect to the ground until it touches the tracks and gets to zero.

1. Inside is a second weighted cylinder

2. Bike moves forward until the crank is at full rearward position

3. No idea

4. wheels from 6oclock to 9oclock

The answer to all four is Danny Devito. Nailed it.

1. Sand

2. Remain there

3. 3 times v1

4. Smoke? xD

2 ) it goes backwards while the pedal seems to go forwards, probably because there is less friction in the wheels than at the pedal.

3 ) (v1 + v2) / 2 = average speed.

not sure about the other two as I haven't given it much thought.

There's is a solution for the third question but you have to consider relativity and cheat a little, let's say the time for the first lap is measured at rest, a friend of yours measure your time for the first lap from the side of the track, he does not move, but you measure your time with your own clock when running for the 2 laps. (that's where I cheat)

Let's say your start slow for your first lap at 90% the speed of light (I know, it's slow but it's just a warm-up) so V1=0.9c

your run the first lap in 1 second (quite a long track, ~270 000 km)

let's assume tm is the time of your clock, moving

tm1=1s

tr is the time of the clock at rest

tr1=1/sqrt(1-V^2/c^2) ~2.29s

you move fast so your clock run slow.

now if you run the second lap so that it last tr1-tm1 (2.29-1=1.29), that mean slowing down to ~70% the speed of light, you will finish the second lap with your clock at 2.29s and comparing to your friend at rest that measured the time you took for the first lap, 2.29s, you will indeed have an average speed that is double the speed of your first lap ! And you didn't even had to go faster on the second one ! Easy peasy lemon squeezy ! đ

Hi every one. Here are my four answers:

1- The Mistry Cylinder: If you hadÂ ever watched the "Curiosity Show", you would have know the answer. Simply, there is a weight hanged in the middleÂ by a rubber spring, which its terminals are fixed at the center of the sides of the cylinder.

2- TheÂ Bicycle:Â It will move "forward" until the pedal moves 90 degree backwardsÂ as if it points to 9 O'clock.

3- The Lap Riddle: Simply, (v1+v2)/2 = 2v1Â

~~—~~> solving "v2" givesÂ Â " v2=3v1 " meaning: you have to run 3 times faster than your first lap speed.4- The Train Riddle: This one is difficult to answer since the whole train moves forward, and it is hard to imagine that something inside the train could move backwardsÂ relative to the ground. However, I have a good guess;Â it isÂ "Â The Wheels" sinceÂ they areÂ the only parts of the train that accelerate and decelerate with respect toÂ both the ground and the train. Also, this video is about rotating objects, thus it is logical to chose "The Wheels" as a guess.

Without looking at the responses already in place: 1) The bottle has honey or a similarly viscus fluid inside that is going to shift slower than gravity wants to move the bottle. 2) The bike will move forward as the string pulls the peddle back and up. This will continue until the peddle is at 90 degrees facing backwards at which point the bike will stop… assuming you didn't pull really hard and then let go. 3) (2)v1 = (v1 + v2) / 2 which reduces to v2 = (3)v1. 4) The wheels of the train. When any wheel travels over a stationary surface half the wheel is moving backwards at any particular moment.

— mike drop —

Q3: in the second lapÂ you have to run infinitely fast. V=â.

1)cylinder had some some obstructions on outer sides.

2)forward

3) not possible

4) smoke

1. Another cylinder with adhesive points. Convenient answer based off knowledge of the environment. He has tape.

2. Move backwards with wheel driving pedals in reverse. Backward velocity of the system greater than instantaneous forward velocity of the pedal. Two systems. Angular velocity of input gear, pedal, less than driven gear, wheel — gear ratio < 1. Additionally, greater torque ground to wheel than string to pedal. Smaller radius of pedal with equal force equates to smaller moment of force driving the system forward relative to force driving the wheel.

3. Overall speed. Lap 2 time equals zero.

Instantaneous velocity. Vâ = 2Vâ

Vâ = (Vâ + Vâ) / 2

4Vâ = Vâ + Vâ

Vâ = 3Vâ

Cheeky velocity. Vâ = 2d / (tâ + tâ)

Vâ =

~~d / tâ –~~run it backwardsVâ = 2Vâ

= -2d / tâ

2d / (tâ + tâ) = -2d / tâ

tâ = -tâ – tâ

tâ = -2tâ

Vâ = d / tâ

= -d / 2tâ

2Vâ = -d / tâ

= Vâ

Vâ = Â˝Vâ

Run lap 2, half the speed, backwards!

4. Ignoring plastic deformation of the track with Newton's third law silliness. Any part of the wheel radius greater than radius point of contact. The part of the wheel below the track head. Logically, point of contact has zero forward velocity. Anything above moves forward relative the train. Anything below must have reverse velocity.

1. My first guess was a viscous liquid in the mystery cylinder, but after watching its behavior closely I don't think that's right. The rolling starts and stops too fast, and with the wobbling it looks more like a freely-moving weight swinging back and forth. So now I have no idea.

2. The bicycle should go forward.

3. I think the mathematical answer is "undefined" because you end up dividing by zero. If you run 100m in 100 seconds, you'd have to move another 100m in zero time in order to cover 200m in 100 seconds, which is what would be required to double the average speed.

4. The part of the train's wheel in contact with the track is always motionless in respect to the ground. The part of it extending lower than the surface of the track on the inside must be moving backward as the wheel rotates.

"…….there is a part of the train which is actually moving BACKWARDS WITH RESPECT TO THE GROUND."

I go with the lower half of the train wheels.

Or maybe the part of the wheel that is 45 degrees left and right of the contact point, which is essensialy the same thing.