Front flip of a bike. Conserving Angular Momentum.

In summary: Can you calculate the time for the bike to travel to the level of the landing point?In summary, the problem involves a dirt bike with a mass of 100 kg and length of 2 m, attached with two wheels of mass 40 kg and radius 0.5 m. The bike goes off a spine ramp at a 1 degree angle to the vertical, and the question is at what time should the rider pull the brakes to do a front flip with minimum takeoff speed. The answer is a real number. The bike lands parallel to the landing ramp with both wheels in contact, and the back wheel touching the top of the ramp. The wheels are treated as rings with all the mass concentrated at the radius, and the
  • #1
Satvik Pandey
591
12

Homework Statement


We model a dirt bike as a rod of mass (M) 100 kg, and length (L) 2 m to which are attached two wheels of mass (m)=40 kg kg, and radius R=0.5 m. The bike goes off a spine ramp with angle ##\phi=1deg.## to the vertical on either side (see diagram below).

Our question is, at what time, ##t_{pull}## (in seconds), after leaving the ramp, should the rider pull the brakes (stopping both wheels instantaneously) if they want to do a front flip with minimum takeoff speed?

  • The answer to this physics question is a real number.
  • On landing, the bike is parallel to the landing ramp, both wheels are in contact with the landing surface, and the back wheel contacts the very top of the ramp.
  • The wheels are rings, i.e. the spokes are massless and all the mass is concentrated at the radius and ##I=mR^{2}##.
  • For simplicity, assume that both wheels lose contact with the ramp at the same angular velocity, ##v/R## , where ##v## is the velocity of the motorcycle upon losing contact with the ramp.
  • The bike coasts off the ramp, i.e. there is no power from the engine putting a torque on the wheel.
  • The bike lands on the opposite side and does a full front flip, I.e. 540 - 2= 538 degrees
Mecc.jpg


Homework Equations

The Attempt at a Solution


[/B]
I first tried to analyse the problem. When the brakes are applied then the wheels stop spinning and as these forces are internal forces, if we consider bike as a system, so they can not brought deviation in the trajectory of the CoM of the system.

As these are internal force so we can conserve angular momentum of the system about its CoM. This will give the magnitude of the angular velocity after the brakes are applied. We can also conserve ttotal energy of the system.

Am I right??

When brakes were not applied till then the bike was not rotating about the its CoM. So the orientation of the bikes remains same.

ssss.png


As the wheels were not rotating about the CoM of the bike before the application of the brakes. So the velocity of the CoM of the wheels wrt the CoM of the system is zero.

Now the angular momentum of the wheels about the CoM( let it be point O) of the frame is equal to
Angular momentum of the CoM of the wheels about point O+Angular momentum of wheels about its CoM.Now the 1st term is zero be cause the wheels were not rotation about point O. So

Initial angular momentum is ## 2I_{w}\omega##.

When the brakes are applied then whole bike would start rotating about the CoM of the bike. Let the angular velocity be ## \omega_{0}##.

So Final Angular momentum is ##I_{system}\omega_{0}##.

So ##I_{system}\omega_{0}=2I_{w}\omega##.

After that I would get a constraint relation that the bike have to rotate 538 deg. in the time laft to land on the ground. I mostly make mistakes in conserving Angular momentum. :confused:

Have I done correctly this time?:oops:

EDIT: I have taken Counter Clockwise as +ve while calculating ##L##.



 
Physics news on Phys.org
  • #2
All looks right so far, except that work will not be conserved when the brakes are applied. But you don't seem to have used that anyway.
 
  • #3
So,

##I_{system}=\frac{ML^2}{12}+2 \left( m{ R }^{ 2 }+m{ \left( \frac { l }{ 2 } \right) }^{ 2 } \right) ##

##I_{w}=mR^{2}##

So on putting these I got

##\omega_{0}=0.3v##

Now time period of projectile is

##\frac { 2v\cos { \phi } }{ g } ##

Time requires by wheels to rotate 538 deg. =## \frac { 269\pi }{ 90(0.3v) } ##

So Answer is ##\frac { 2v\cos { \phi } }{ g }- \frac { 269\pi }{ 90(0.3v) } ##.

I think I need the value of 'v'. How should I get it?
 
  • #4
Satvik Pandey said:
Time requires by wheels to rotate 538 deg. =## \frac { 269\pi }{ 90(0.3v) } ##

So Answer is ##\frac { 2v\cos { \phi } }{ g }- \frac { 269\pi }{ 90(0.3v) } ##.

I think I need the value of 'v'. How should I get it?
You mean, the time required for the whole bike to rotate, yes?
It isn't made clear, but I assume the start point is when the front wheel leaves the ramp. That makes it symmetric with the landing position.
Consider the horizontal morion while airborne.
 

Related to Front flip of a bike. Conserving Angular Momentum.

1. What is a front flip of a bike?

A front flip of a bike is a trick performed on a bicycle where the rider rotates the bike around its front wheel while in the air, landing back on both wheels.

2. How is angular momentum conserved during a front flip of a bike?

Angular momentum is conserved during a front flip of a bike through the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant as long as there are no external torques acting on the system. In a front flip, the rider initiates the rotation by pushing down on the handlebars, creating an angular momentum which is then conserved as the rider and bike rotate in the air.

3. What factors affect the front flip of a bike?

The front flip of a bike can be affected by several factors, including the speed and angle of approach, the rider's body position and technique, and the weight and design of the bike. These factors can impact the amount of angular momentum generated and the stability and control of the bike during the flip.

4. Is a front flip of a bike dangerous?

Like any extreme sport, there is a risk of injury associated with performing a front flip of a bike. However, with proper training and safety precautions, the risk can be minimized. It is important for riders to have a solid understanding of the physics and technique involved in order to perform the trick safely.

5. Can a front flip of a bike be performed on any type of bike?

A front flip of a bike can be performed on most types of bicycles, including BMX bikes, mountain bikes, and even road bikes. However, the design and weight of the bike may affect the difficulty and stability of the trick. Riders should choose a bike that is suitable for their skill level and the type of terrain they will be performing the front flip on.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
957
  • Introductory Physics Homework Help
Replies
17
Views
182
  • Introductory Physics Homework Help
2
Replies
45
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
832
  • Introductory Physics Homework Help
10
Replies
335
Views
8K
  • Introductory Physics Homework Help
Replies
5
Views
176
  • Introductory Physics Homework Help
Replies
30
Views
2K
Replies
9
Views
3K
Back
Top