Applying a force in a wheel bicycle

  • #1

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:
F=mat=mrα→α=F/mr→α=10/(1,5·0,33)=20,2 rad/s2

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/s


Is 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?
 

Answers and Replies

  • #2
haruspex
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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?
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.
 
  • #3
[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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