Understanding Bike Mechanics: Simplifying the Tailwhip with Bar and Wheel Pivots

  • Thread starter Thread starter joshd
  • Start date Start date
  • Tags Tags
    Bike Mechanics
AI Thread Summary
The discussion focuses on the mechanics of performing a tailwhip on a bike, particularly how the wheel's radius affects the torque required to spin it around the pivot at the headtube. It explores the concept of simplifying the wheel and bar as pivot points, questioning whether a smaller wheel radius decreases the force needed for spinning. The conversation highlights the conflicting torque effects of the wheel's back and front edges in relation to the pivot. Additionally, it considers the implications of switching wheel sizes while maintaining mass, emphasizing that the distances from the pivot do not balance out. Understanding these dynamics is crucial for grasping the mechanics involved in executing a tailwhip effectively.
joshd
Messages
24
Reaction score
0
I was discussing this with a friend:

Say you are doing a tailwhip on a bike (bike spins around pivot at headtube)

Simpify this by looking at it as a bar pivoted at one end, with a wheel mounted to the bar with the radius in the direction of the bar. We want to spin this around the pivot in the direction of the wheels axle. (The wheel is spinning, but I am not sure if this makes any difference?)

Now, if we make the wheel's radius smaller, does the force required to spin the wheel/bar around the pivot decrease? The "back edge" of the wheel is closer to the pivot, so less torque is needed to turn it around the pivot, but the "front edge" of the wheel is further from it, so more torque is needed to turn it around the pivot?

However, can the wheel not be though of a point mass, positioned at the axle? In this case, provided the wheel's mass does not change, the radius of the wheel is independent?

Thanks.
 
Physics news on Phys.org
I did not probably get it right (the movement part, I'm not really a bike monster, so I'm not sure about what a tailwhip is), but it looks like a spinning wheel doing O -> | -> O etc, and a bar doing something like \ -> | -> / with an overall movement like this one (3 photographs :P)
\ - | - /
O | O
watching from the side of the bike.
At this point you should have the torque of the wheel and the one of the bar. Reducing thw wheel's radius should reduce its torque leaving the same torque to the bar that remains the same as before.
So i didn't get the whole
The "back edge" of the wheel is closer to the pivot, so less torque is needed to turn it around the pivot, but the "front edge" of the wheel is further from it, so more torque is needed to turn it around the pivot?
thing.

At this point you should be able to see whether the change of radius is influent, depending on the density of mass of the wheel and the one of the bar, to see if the wheel influences the torque enough to make something change effectively.

Hope that helps.

SS
 
Here:

What the question is aiming at is this: what if you switch your wheels for 24" dia. instead of 26"? Assume mass of wheel stays the same.

The axle of the wheel is at the same position, so the wheel's mass is centred at the same point. But, the "back edge" of the wheel is 1" closer to the headtube (the pivot), and the "leading edge" of the wheel is 1 inch further away. But these effects don't balance, since they are different distances from the pivot?
 
Last edited by a moderator:
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top