# Applying a force in a wheel bicycle

• Guillem_dlc
In summary: IαThis formula is used when the axis of rotation is known, that is, the mass is fixed and the axis of rotation varies.[QUOTE="haruspex, post: 6059313, member: 334404"]To find the pair of the net force is done by means of the following formula:You mean the torque, not the pair.Rather simpler would be to multiply the force by the "lever arm", i.e. the radius of the wheel.[QUOTE="haruspex, post: 6059313, member: 334404"]Yes, I mean the torque, thank you for pointing that out. And that is a good suggestion to simplify the calculation. Thank you
Guillem_dlc

## Homework Statement

We have a bicycle and lift the front wheel. And with the hand we apply a force of 10 N down and we turn the wheel. The wheel is 26 inches in diameter and weighs 1.5 kg. Find the angular acceleration, the torque of the net force and the angular velocity after a lap.

## Homework Equations

r=d/2; F=ma_t=mrα; Γ=mr2α...

## The Attempt at a Solution

Data:
vector(F)=10 N
d=26''
m=1,5 kg

The first step is to pass the data into units of the international system (S.I), and then look for the radius of the wheel. The only data that is not in units of S.I is the diameter, therefore
26 · (0,0254 m)(1) = 0,66 m
r=d/2 → r=(0,66)/2=0,33 m

Once we are going to calculate the angular acceleration. Based on the formula of Newton's second law and since the force we exercise already acts in the tangential direction we isolate and replace:

To find the pair of the net force is done by means of the following formula:
Γnet=mr2α→Γnet=1,5·0,332·20,2=3,3 N·m

And finally we calculate the angular velocity after a turn. One turn is 360 degrees which radiant is equivalent to 2π rad
ω202+2αφ→ω=sqrt(2αφ)→ω=sqrt(2·20,2·2π)=15,93 rad/sIs this exercise well resolved? What could be added so that it was more physically solved? And any idea to include in the exercise or modify it?

Guillem_dlc said:
F=mat=mrα→α=F/mr
This is correct, but it is unclear how you arrived at it. What if the wheel were a uniform disc?
Guillem_dlc said:
To find the pair of the net force is done by means of the following formula:
You mean the torque, not the pair.
Rather simpler would be to multiply the force by the "lever arm", i.e. the radius of the wheel.

[QUOTE="haruspex, post: 6059313, member: 334404"This is correct, but it is unclear how you arrived at it. What if the wheel were a uniform disc?[/QUOTE]
Before I do the proof of the formula.

To demonstrate the equation of Newton's second law to rotation, we will base ourselves on a drawing.
Formed by a mass particle m attached to a rigid bar of length r and too insignificant and to a rotation axis. Therefore the particle moves in a circle of radius r. On the particle a force vector(F). is applied. It only works in the tangential direction Ft. With the second law of Newton you get it
Ft=mat
where Ft=F·sinφ and at is the tangential acceleration. We have that at=rα where r is the rigid bar, which is equivalent to the radius of the circumference and α is the angular acceleration, therefore:
Ft=mrα
We know that the moment or torque Γ=rFt and to be able to replace we have to multiply the radius r on each side
rFt=mr2α → Γ=mr2α
A rigid object that rotates on an axis (the wheel of the bicycle) is simply a set of particles that are forced to move in a circular and at the same speed. Applying the equation prior to the particle and the set, it remains
Γi net=miri2α
where Γi net is the torque of the net force on particle i. If we add the rest of particles, the terms of the following expressions are obtained
∑Γi net=Σmiri2α=(∑miri2
and if we replace the moment of inertia, but we know that I=Σmiri2, it is:
∑Γi net=Iα
and taking into account that the sum of forces of the particles is equivalent to the net external force that acts on the whole set, the formula of the second Newton law is obtained at the rotation:
Γnet ext=Iα

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## 1. How does applying a force in a wheel bicycle make it move?

When a force is applied to the pedals of a wheel bicycle, it causes the crank to rotate. This rotation is then transferred to the chain and the rear wheel, causing it to turn and propel the bicycle forward.

## 2. How much force is needed to move a wheel bicycle?

The amount of force needed to move a wheel bicycle depends on factors such as the weight of the rider, the terrain, and the resistance of the bicycle's components. Generally, a moderate amount of force is needed to move a bicycle at a comfortable speed.

## 3. Can applying too much force damage a wheel bicycle?

Yes, applying too much force can potentially damage a wheel bicycle. This is especially true if the force is applied abruptly or unevenly, causing stress on the bicycle's components. It is important to pedal smoothly and avoid sudden, excessive force.

## 4. How does the size of the wheel affect the force needed to move a bicycle?

The size of the wheel does not directly affect the force needed to move a bicycle. However, larger wheels have a greater circumference, which means they cover more distance with each rotation. This can make it easier to maintain a certain speed with less force, but it also requires more force to initially get the bicycle moving.

## 5. Can the direction of the force affect the movement of a wheel bicycle?

Yes, the direction of the force can affect the movement of a wheel bicycle. When the force is applied perpendicular to the ground, it creates a torque that causes the wheel to rotate. However, if the force is applied in a direction that does not align with the axle of the wheel, it can cause the bicycle to veer off course or even tip over.

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