https://isaacphysics.org/questions/monster_truck_front_flip?board=d44d9938-9677-46d4-83dc-58b765b19651&stage=further_a

Hello, I tried representing all the important stuff with the equations below but I must be either missing something or the equations are wrong.. or both.

If someone could point me in the right direction, it would be greatly appreciated.

Hello, I tried representing all the important stuff with the equations below but I must be either missing something or the equations are wrong.. or both.

If someone could point me in the right direction, it would be greatly appreciated.

Diagram:

Working:

I said w = 7/3 pi since it does two 30 degree rotations plus a full turn - so 420 degrees in total but tbh im not sure if this is correct.

edit: sorry I made a mistake for my equation of torque - it should be delta_w/1.5s not delta_w/delta_t but even still I cant really see a way forward.

Working:

I said w = 7/3 pi since it does two 30 degree rotations plus a full turn - so 420 degrees in total but tbh im not sure if this is correct.

edit: sorry I made a mistake for my equation of torque - it should be delta_w/1.5s not delta_w/delta_t but even still I cant really see a way forward.

(edited 1 month ago)

Original post by mosaurlodon

Diagram:

Working:

I said w = 7/3 pi since it does two 30 degree rotations plus a full turn - so 420 degrees in total but tbh im not sure if this is correct.

edit: sorry I made a mistake for my equation of torque - it should be delta_w/1.5s not delta_w/delta_t but even still I cant really see a way forward.

Working:

I said w = 7/3 pi since it does two 30 degree rotations plus a full turn - so 420 degrees in total but tbh im not sure if this is correct.

edit: sorry I made a mistake for my equation of torque - it should be delta_w/1.5s not delta_w/delta_t but even still I cant really see a way forward.

The key thing is to get the normal impulse (youve got the initial vertical speed) which gives the required vertical motion, then use that to get the corresponding angular impulse (hint 4) and equate that to the desired change in angular momentum, Iw, where w is the rotational speed.

All are relatively simple calculations, so try and write down the concept youre using clearly and what youre trying to find.

im pretty sure the first half of this is correct but the second half im not too sure about.

So initially the truck has 0 upwards velocity and v_h1 horizontal velocity and since the impulse acts at theta to the vertical then the upwards velocity increases to u while the horizontal velocity decreases to v_h2.

but I dont really see how to work from here

So initially the truck has 0 upwards velocity and v_h1 horizontal velocity and since the impulse acts at theta to the vertical then the upwards velocity increases to u while the horizontal velocity decreases to v_h2.

but I dont really see how to work from here

(edited 1 month ago)

Original post by mosaurlodon

im pretty sure the first half of this is correct but the second half im not too sure about.

So initially the truck has 0 upwards velocity and v_h1 horizontal velocity and since the impulse acts at theta to the vertical then the upwards velocity increases to u while the horizontal velocity decreases to v_h2.

but I dont really see how to work from here

So initially the truck has 0 upwards velocity and v_h1 horizontal velocity and since the impulse acts at theta to the vertical then the upwards velocity increases to u while the horizontal velocity decreases to v_h2.

but I dont really see how to work from here

Wbf, that may mean something to you but its not easy for another person to understand it.

1) N Delta t

is the linear impulse and if you can get that, then you can get I as you know r=1.5 and 3w/2 = 7pi/3. Youve not written down that last part?

2) N is perpendicular to the surface and you know (can calculate) what the vertical impulse should be? So a bit of simple trig to get N, then the previous paragraph.

3) N does affect the horizontal motion, but there is no reason/nothing in the question to say its important so why bring it into the analysis?

Your diagram with N in #1 is correct, but youve not sketched the things that are important to get the solution, so determine the vetical impulse, use that to get N, then use that to get I by looking at rotational motion.

Oh this question is looking so much more obvious in hindsight.

Sorry about the horrible working out btw that tends to happen when I get confused - counter productive I guess?

For some reason, I kept looking at N Delta t as a whole rather than vertically which meant I was looking at Delta p as a whole (including the horizontal component) which just spiralled out of hand.

Thank you for your patience

Sorry about the horrible working out btw that tends to happen when I get confused - counter productive I guess?

For some reason, I kept looking at N Delta t as a whole rather than vertically which meant I was looking at Delta p as a whole (including the horizontal component) which just spiralled out of hand.

Thank you for your patience

Actually I do have one more question if you dont mind.

the equation F = Delta p/Delta t is only applicable when F is a resultant (in this case vertical) force so why is the equation not

(Ncos(theta) - mg) = Delta p/Delta t

I get why for the second equation it's just 1.5*N *Delta t = I* Delta w since mg acts on the com which doesnt provide a torque.

the equation F = Delta p/Delta t is only applicable when F is a resultant (in this case vertical) force so why is the equation not

(Ncos(theta) - mg) = Delta p/Delta t

I get why for the second equation it's just 1.5*N *Delta t = I* Delta w since mg acts on the com which doesnt provide a torque.

(edited 1 month ago)

Original post by mosaurlodon

Actually I do have one more question if you dont mind.

the equation F = Delta p/Delta t is only applicable when F is a resultant (in this case vertical) force so why is the equation not

(Ncos(theta) - mg) = Delta p/Delta t

I get why for the second equation it's just 1.5*N *Delta t = I* Delta w since mg acts on the com which doesnt provide a torque.

the equation F = Delta p/Delta t is only applicable when F is a resultant (in this case vertical) force so why is the equation not

(Ncos(theta) - mg) = Delta p/Delta t

I get why for the second equation it's just 1.5*N *Delta t = I* Delta w since mg acts on the com which doesnt provide a torque.

I guess strictly you should as you really want to work out

N Delta t = Delta p + mg Delta t

Probably three reasons

1) Impulses are generally (not always) very short duration events. Theyre used to represent what happens when you have a jump in velocity, so a car crashing or ... and its messy to model what happens (forces, deformation, ...) in the short time when v "jumps" to zero or ... Here, its a similar case so the rear wheels hit the ramp and the impulse is a short duration event. If Delta t ~ 0, then the weight can be neglected. Its a bit like saying no work is done by a force if the object doesnt move.

2) Theres nothing in the question which allows you to model it (so no duration or ...). Not a great reason admittedly but cest la vie.

3) Point 3 says the vertical impulse comes solely from N

(edited 1 month ago)

I see so they made the point 3 assumption because Delta t ~ 0 so the difference between Ncos(theta) and mg would be minimal and since N >> mg then this difference is pretty negligible.

(edited 1 month ago)

Original post by mosaurlodon

I see so they made the point 3 assumption because Delta t ~ 0 so the difference between Ncos(theta) and mg would be minimal and since N >> mg then this difference is pretty negligible.

Again, sleeping on it helps. The previous stuff about short duration events (impulses) isnt the whole story here and the part 3 hint isnt that accurate here. It helps to think about the wheelie the truck is pulling beforehand. In this case there is the usual normal reaction force on the wheel and a forwards force on the wheel driving it forwards. The combination of these two forces passes through the com which means the wheelie is maintained - the zero moment point of the truck is on the base of wheel. When it hits the ramp, this force on the bottom of the wheel still exists in addition to the new impulse at another point on the wheel at 30 degrees to the vertical.

https://media.wired.com/photos/5e2b7e31ecdb900009b53954/master/w_1600,c_limit/Science_frontflipforce.jpg

https://www.wired.com/story/monster-truck-front-flip/

The original "wheelie force" at the bottom of the wheel will still account for the weight of the truck (roughly, as it could be argued that the new force would now account for a bit under half of it) and the new ramp impulse will cause the vertical velocity jump. The difference (sum) in the horizontal components of these forces will mean that the wheelie horizontal force (velocity jump) is reduced so the truck starts to rotate. The reality is that the vertical component of the impulse must account for a bit less than half the weight, but as in the previous post, if the duration is small, then this will be small compared to the impulse component necessary to cause the vertical velocity jump.

Edit - it would have been an interesting part a) to talk about the wheelie phase and then part b) for this question part, as it could have talked about the impact event in a bit more detail. For the question as it is, they just want you to think about what impulse is necessary to give a vertical velocity jump and ignore all this.

(edited 1 month ago)

The fact that there is an entire article explaining the physics of this jump is pretty convenient.

The diagram actually makes some intuitive sense - if the ramp were a lot steeper then the truck would do a lot more flips since the horizontal impulse>>horizontal "F_b" but would have a shorter jump height and I guess if you picture the horizontal of "F_b" to be equal to the horizontal impulse then there wont be a torque and the horizontal velocity will remain constant so the truck will just "jump over" but wont do any flips.

A bit of a tangent but I dont really understand why they say F_b is made up of the Normal Force acting perpendicular and friction? I remember asking my physics teacher about what force pushes cars forward and they said it was newtons third law - similar to how people swim by pushing water - i.e. the normal force acts both perpendicular (to keep the car from falling into the ground) and parallel (to push it forward).

The diagram actually makes some intuitive sense - if the ramp were a lot steeper then the truck would do a lot more flips since the horizontal impulse>>horizontal "F_b" but would have a shorter jump height and I guess if you picture the horizontal of "F_b" to be equal to the horizontal impulse then there wont be a torque and the horizontal velocity will remain constant so the truck will just "jump over" but wont do any flips.

A bit of a tangent but I dont really understand why they say F_b is made up of the Normal Force acting perpendicular and friction? I remember asking my physics teacher about what force pushes cars forward and they said it was newtons third law - similar to how people swim by pushing water - i.e. the normal force acts both perpendicular (to keep the car from falling into the ground) and parallel (to push it forward).

(edited 1 month ago)

Original post by mosaurlodon

The fact that there is an entire article explaining the physics of this jump is pretty convenient.

The diagram actually makes some intuitive sense - if the ramp were a lot steeper then the truck would do a lot more flips since the horizontal impulse>>horizontal "F_b" but would have a shorter jump height and I guess if you picture the horizontal of "F_b" to be equal to the horizontal impulse then there wont be a torque and the horizontal velocity will remain constant so the truck will just "jump over" but wont do any flips.

A bit of a tangent but I dont really understand why they say F_b is made up of the Normal Force acting perpendicular and friction? I remember asking my physics teacher about what force pushes cars forward and they said it was newtons third law - similar to how people swim by pushing water - i.e. the normal force acts both perpendicular (to keep the car from falling into the ground) and parallel (to push it forward).

The diagram actually makes some intuitive sense - if the ramp were a lot steeper then the truck would do a lot more flips since the horizontal impulse>>horizontal "F_b" but would have a shorter jump height and I guess if you picture the horizontal of "F_b" to be equal to the horizontal impulse then there wont be a torque and the horizontal velocity will remain constant so the truck will just "jump over" but wont do any flips.

A bit of a tangent but I dont really understand why they say F_b is made up of the Normal Force acting perpendicular and friction? I remember asking my physics teacher about what force pushes cars forward and they said it was newtons third law - similar to how people swim by pushing water - i.e. the normal force acts both perpendicular (to keep the car from falling into the ground) and parallel (to push it forward).

There are a few explanations about the dynamics so forces and torques (just google them) but its probably easier just to imagine the movement of a rolling wheel

https://en.wikipedia.org/wiki/Rolling#/media/File:Rolling_animation.gif

So a point on the wheel traces out a cycloid

https://en.wikipedia.org/wiki/Cycloid

This can only occur if the linear force T/r which due to the torque T at the center of a wheel of radius r is equal to the surface friciton at the point of contact and the relevant point on the wheel is instantaneously stationary.

(edited 1 month ago)

Original post by mosaurlodon

The fact that there is an entire article explaining the physics of this jump is pretty convenient.

The diagram actually makes some intuitive sense - if the ramp were a lot steeper then the truck would do a lot more flips since the horizontal impulse>>horizontal "F_b" but would have a shorter jump height and I guess if you picture the horizontal of "F_b" to be equal to the horizontal impulse then there wont be a torque and the horizontal velocity will remain constant so the truck will just "jump over" but wont do any flips.

A bit of a tangent but I dont really understand why they say F_b is made up of the Normal Force acting perpendicular and friction? I remember asking my physics teacher about what force pushes cars forward and they said it was newtons third law - similar to how people swim by pushing water - i.e. the normal force acts both perpendicular (to keep the car from falling into the ground) and parallel (to push it forward).

The diagram actually makes some intuitive sense - if the ramp were a lot steeper then the truck would do a lot more flips since the horizontal impulse>>horizontal "F_b" but would have a shorter jump height and I guess if you picture the horizontal of "F_b" to be equal to the horizontal impulse then there wont be a torque and the horizontal velocity will remain constant so the truck will just "jump over" but wont do any flips.

A bit of a tangent but I dont really understand why they say F_b is made up of the Normal Force acting perpendicular and friction? I remember asking my physics teacher about what force pushes cars forward and they said it was newtons third law - similar to how people swim by pushing water - i.e. the normal force acts both perpendicular (to keep the car from falling into the ground) and parallel (to push it forward).

After a bit of googling, this yoyo example is a good explanation where the torque comes from pulling the string

https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-12-rotations-and-translation-rolling/36-4-worked-example-yoyo-pulled-along-the-ground/

The videos before this one explain some of the concepts a bit more but its reasonably clear.

The links are very useful thanks - for the yoyo one im guessing he writes I_cm since the yoyo doesnt have uniform density otherwise im not sure why he doesnt simplify further since I think it should just be a function of m and r.

Original post by mosaurlodon

The links are very useful thanks - for the yoyo one im guessing he writes I_cm since the yoyo doesnt have uniform density otherwise im not sure why he doesnt simplify further since I think it should just be a function of m and r.

Pretty much as you say, I guess they just wanted to keep it general so it could refer to a uniform density disc or a ring where the mass is on the rim or ...

I emailed Isaac about this question before hand and heres the reply if youre curious - I asked them the same question of why you should not consider weight for the resultant force.

"You are right to consider the effect of the constant weight. In the impulse limit of delta t tending to zero, the applied force F tends to infinity, so that the product of the two is a finite result. So with a very large force for a very short time, subtracting the weight for the large force becomes a negligible change. So it does require a short time for the impulse to act.

It is not very intuitive, but it is a result of taking a limit."

They dont mention normal reaction force (or the "wheelie force") - so I guess it would be a combination of both arguments of why weight can be ignored, but tbh the wheelie force argument makes more sense to me.

"You are right to consider the effect of the constant weight. In the impulse limit of delta t tending to zero, the applied force F tends to infinity, so that the product of the two is a finite result. So with a very large force for a very short time, subtracting the weight for the large force becomes a negligible change. So it does require a short time for the impulse to act.

It is not very intuitive, but it is a result of taking a limit."

They dont mention normal reaction force (or the "wheelie force") - so I guess it would be a combination of both arguments of why weight can be ignored, but tbh the wheelie force argument makes more sense to me.

(edited 1 month ago)

Original post by mosaurlodon

I emailed Isaac about this question before hand and heres the reply if youre curious - I asked them the same question of why you should not consider weight for the resultant force.

"You are right to consider the effect of the constant weight. In the impulse limit of delta t tending to zero, the applied force F tends to infinity, so that the product of the two is a finite result. So with a very large force for a very short time, subtracting the weight for the large force becomes a negligible change. So it does require a short time for the impulse to act.

It is not very intuitive, but it is a result of taking a limit."

They dont mention normal reaction force (or the "wheelie force") - so I guess it would be a combination of both arguments of why weight can be ignored.

"You are right to consider the effect of the constant weight. In the impulse limit of delta t tending to zero, the applied force F tends to infinity, so that the product of the two is a finite result. So with a very large force for a very short time, subtracting the weight for the large force becomes a negligible change. So it does require a short time for the impulse to act.

It is not very intuitive, but it is a result of taking a limit."

They dont mention normal reaction force (or the "wheelie force") - so I guess it would be a combination of both arguments of why weight can be ignored.

Pretty much agree with that. On average, the reaction at the bottom of the wheel would account for a bit over half the weight and the reaction on the ramp a bit less than a half, though in the limit as the duration -> 0, this is small in comparison to the impulse necessary to cause the velocity jump. So it becomes a bit of an argument about the actual duration of the event and point 3 sort of covers that though it would have been more accurate to say all the impulse causes the velocity jump (rather than the other way round) or that the duration was small so that weight could be neglected. Though with tyres as large as they are, the duration of the impulse is probably not infinitesimally small.

(edited 1 month ago)

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