Bicycle problem?

Question

Evaluate how fast can a bicycle ride that after pushing a stick into the front wheel, the bicyclist won't flip over his head? Wheels and a stick are strong and unbreakable.
(I need to create a physical model with variables, equations and so on).

Thanks in advance.

Answer 1

Most likely the rider will flip over under real conditions. Your assumption of unbreakable wheels and stick guarantee it.

Look at a FBD of the front wheel first:
The forward momentum of the bike will be resisted only by the friction of the front wheel and asphalt. Assume a coefficient of friction of about 0.6 to 0.8.

Then look at the moment of inertia of the bike and rider. Since the front wheel will slip very little if at all, that force becomes a torque on the system. All of the forward momentum of the bike and rider will be converted to angular momentum. Then gravity will resist until the center of mass passes above the top of the rotation - the only hope the rider has is that the forward speed was low enough that he stops before passing through apogee. If not, gravity helps slam his head into the pavement - ouch.

To get more quantitative, let's model this with some equations:

The first approximation is that all of the rotation will occur about the axel of the front wheel as the center. The bicycle and rider will be a point mass of 90 kg on a lever 100 cm from the axel and at an angle of -pi/4 radians from the vertical. The contact patch of the road and tire will be a torque that is at pi radians from the vertical and constant at that position. The magnitude of the torque will be considered in a moment (no pun intended).

Assume that all of the forward momentum of the mass gets converted to angular momentum in an instant.

Therefore,
m*v=I*w
the I=m*R^2
where R=100 cm or 1 meter
therefore
v=R^2*w
or
w=v/R^2
since R=1
w=v

Assuming conservation of energy once the angular motion starts, there is a gain in potential energy as the rider rotates the fron wheel to the point where the R is vertical. BTW the radius of the front wheel is .35 meters.
The center of mass starts at a height of
.35+R*cos(pi/4), or .35+sqrt(2)/2. and when vertical is at an angle of
.35 +R, or 1.35 m

The gain in potential energy is
90*g*(1.35-.35-sqrt(2)/2)
or
45*g*(2-sqrt(2))
using g=9.81
this is
258.6 J

This is equal to the loss of kinetic energy of the system
or
so if the rider reaches the vertical position, then all of the energy gets converted
258.6=.5*90*R^2*w1^2
from above
258.6=.5*90*v^2
solve for v
2.4 m/s or
8.64 km/hr That's not so fast, so any speed greater than, 8.64 km/hr and the rider rotates above vertical and lands on their head. Under this condition, the rider will lose
.35+sqrt(2)/2 of height, which will be added to the angular kinetic energy when slamming into the ground.
Or, .5*I*v^2+m*g*(.35+sqrt(2)/2)=.5*I*w2^2
Again, I=m*R^2, R=1, so
v^2+2*g*(.35+sqrt(2)/2))=w2^2
translational speed relates to angular as w*r=v
in our case, R=1. so the loss in PE gets converted to the new speed of impact as
v2=sqrt(v^2+20.74)

as long as v>2.4 m/s

If the rider is traveling at a speed of 21.6 km/hr, or 6 m/s
the speed at impact is 7.53 m/s or 27 km/hr

There are some inaccuracies in this solution. The most dramatic is that the rider and bike retain some tranlation forward momentum since the rotation is eliptical, not circular as I modeled.

The center of rotation is actually the contact patch between the road and front tire. This is one focus of the ellipse of motion (I also assumed the whole bike and rider remain rigid through the crash). I won't derive this solution since I believe that the loss in accuracy is small since the total momentum is conserved, and the rotatinal momentum will dominate.

Lastly, let's look at the torque on the front wheel from the contact with the road. Since the rider and bike will decelerate as the momentum shifts from translational to rotational, and the rider rises to lose PE, there is a deceleration. This can be assumed to be translated into an average torque at the point of contact, such that
258.6 J is the loss in KE in the time it takes the rider to rise.


The faster the speed of the rider, the faster the rise, the greater the impulse on the front wheel.
At an average of 5 m/s (about 6 m/s initial v), the rise takes
0.157 seconds.

j

Answer 2

First, we need to assume that the friction between the tire and the road is sufficient to prevent skidding. If skidding is a possibility then the problem gets more complicated.

When the stick enters the wheel it stops rotating, and the question is whether or not the bike and rider will rotate about the center of the front wheel. To simplify the problem you need to define where the combined center of gravity (CG) of you and your bike is with respect to the center of the front wheel (the Pivot Point, PP). X could be the distance of the CG behind the PP, and Y could be the distance above the PP. In order for you to completely flip over the handlebars you would have to have enough kinetic energy to rotate the center of gravity over the top of the PP, at which point you would fall by gravity the rest of the way over. Rotating the center of gravity over the top of the PP increases the potential energy of you and your bike as the CG moves higher during the rotation. To calculate how much the CG rises we need to know how far the CG is from the PP. We can get that by using Pythagorean's Theorem.

D^2 = X^2 + Y^2 (D is the linear distance from the CG to the PP)

The increase in height of the CG is:
H = D - Y

So the conservation of energy equation is (change in Kinetic Energy = change in Potential Energy):
KE = PE
1/2 x m x V^2 = m x g x H

Where m is the total mass of you and the bike and V is the initial Velocity of the bike. Now you can see how the location of the CG effects the stability of the bike.