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Excerpt
hiddentrue

Determine how the weight of a bicycle plus rider is divided between the wheels in various circumstances.

Composition Setup
Image Added
 Deck of Cards
idbigdeck
 Wiki Markup
{excerpt:hidden=true}Determine how the weight of a bicycle plus rider is divided between the wheels in various circumstances.{excerpt}

{table:align=right|cellspacing=0|cellpadding=1|border=1|frame=box|width=40%}
{tr}
{td:align=center|bgcolor=#F2F2F2}{*}[Examples from Rotation]*
{td}
{tr}
{tr}
{td}
{pagetree:root=Examples from Rotation}
{search-box}
{td}
{tr}
{table}

{composition-setup}{composition-setup}

!bike1.png!

{deck:id=bigdeck}
{card:label=Part A}

h2. Part A

A cyclist sits on their bike at rest, supported only by the two wheels of the bike.  The combined mass of the bike plus cyclist is 95 kg.  The center of mass of the bike plus cyclist is 0.75 m above the ground, 0.42 m forward of the center of the rear wheel and 0.66 m behind the center of the front wheel.  What is the size of the normal force exerted by the ground on each wheel?

h4. Solution

{toggle-cloak:id=sysa} *System:* {cloak:id=sysa} Cyclist plus bicycle are treated as a single [rigid body].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External forces from the earth (gravity) and the ground (normal force and friction).{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:* 

{cloak:id=appa} 

{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diaga}
We first sketch the situation and construct a coordinate system.

!bikeparta.jpg!

{cloak:diaga}

{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

{cloak:id=matha}

We now write the equations of [Newton's 2nd Law|Newton's Second Law] for the center of mass and of torque balance for the system.  Since the system is at rest, we can put the axis wherever we choose.  We select the point of contact of the rear wheel with the ground to remove an unknown from the torque equation.  

{latex}
 Card
labelPart A
Part A
A cyclist sits on their bike at rest, supported only by the two wheels of the bike. The combined mass of the bike plus cyclist is 95 kg. The center of mass of the bike plus cyclist is 0.75 m above the ground, 0.42 m forward of the center of the rear wheel and 0.66 m behind the center of the front wheel. What is the size of the normal force exerted by the ground on each wheel?
Solution
 Toggle Cloak
idsysa
 System: 
 Cloak
idsysa
 Cyclist plus bicycle are treated as a single .
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
External forces from the earth (gravity) and the ground (normal force and friction).
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
 and .
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
  Diagrammatic Representation 
 Cloak
iddiaga
We first sketch the situation and construct a coordinate system.
Image Added
 Cloak
diaga
diaga
 Toggle Cloak
idmatha
  Mathematical Representation 
 Cloak
idmatha
We now write the equations of Newton's 2nd Law for the center of mass and of torque balance for the system. Since the system is at rest, we can put the axis wherever we choose. We select the point of contact of the rear wheel with the ground to remove an unknown from the torque equation. 
 Latex
\begin{large} \[ \sum F_{y} = - mg + N_{\rm front} + N_{\rm rear} = 0 \]
\[\sum \tau = N_{\rm front} (L_{\rm front}+L_{\rm rear}) - mg L_{\rm rear}  = 0\]\end{large}
{latex}

{note}
 Note
Newton's 
 
 Law 
 
 for 
 
 the 
 _
x
_ 
 direction 
 
 is 
 
 trivial 
 
 (0 
 
 = 
 
 0), 
 
 so 
 
 we 
 
 have 
 
 ignored 
 
 it.
{note}


The 
 
 torque 
 
 equation 
 
 immediately 
 
 gives:


{
 Latex
}
\begin{large}\[ N_{\rm front} = \frac{mgL_{\rm rear}}{L_{\rm front}+L_{\rm rear}} = \mbox{362 N}\]\end{large}
{latex}


Force balance in the _y_ direction then gives:

{latex}
Force balance in the y direction then gives:
 Latex
\begin{large}\[ N_{\rm rear} = mg - N_{\rm front} = \frac{mgL_{\rm front}}{L_{\rm front}+L_{\rm rear}} = \mbox{569 N}\]\end{large}
{latex}

{cloak:matha}
{cloak:appa}
{card}
{card:label=Part B}

h2. Part B


Suppose the cyclist described in Part A is coasting along at a constant speed when suddenly a car pulls out in front of them.  The cyclist hits the brakes and locks both wheels (they stop rotating and begin to skid).  The bike skids straight forward.  If the cyclist and bike are decelerating at 0.55 _g_, what is the size of the normal force exerted by the ground on each wheel? 

h4. Solution

{toggle-cloak:id=sysb} *System:* {cloak:id=sysb} Cyclist plus bicycle are treated as a single [rigid body].{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}External forces from the earth (gravity) and the ground (normal force and friction).{cloak}

{toggle-cloak:id=modb} *Model:*  {cloak:id=modb}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appb} *Approach:*  

{cloak:id=appb}
{toggle-cloak:id=diagb} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagb}
We again sketch the situation and construct a coordinate system.

!bikepartb.jpg!

{cloak:diagb}

{toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color}

{cloak:id=mathb}

We now write the equations of [Newton's 2nd Law|Newton's Second Law] for the center of mass and of torque balance about the center of mass for the bicycle.  

{note}The bicycle's center of mass is accelerating linearly in the negative _x_ direction, but the bicycle is not rotating about its center of mass.  Thus the torques about the center of mass must balance.{note}

{warning}Because the bicycle is accelerating, the only point moving with the bike that can be chosen as an axis of rotation is the center of mass.  Try repeating the problem with another axis (such as the point of contact of the front wheel with the ground) and you should find that the torques do _not_ balance).{warning}

{latex}
 Cloak
matha
matha
 Cloak
appa
appa
 Card
labelPart B
Part B
Suppose the cyclist described in Part A is coasting along at a constant speed when suddenly a car pulls out in front of them. The cyclist hits the brakes and locks both wheels (they stop rotating and begin to skid). The bike skids straight forward. If the cyclist and bike are decelerating at 0.55 g, what is the size of the normal force exerted by the ground on each wheel? 
Solution
 Toggle Cloak
idsysb
 System: 
 Cloak
idsysb
 Cyclist plus bicycle are treated as a single .
 Toggle Cloak
idintb
 Interactions: 
 Cloak
idintb
External forces from the earth (gravity) and the ground (normal force and friction).
 Toggle Cloak
idmodb
 Model: 
 Cloak
idmodb
 and .
 Toggle Cloak
idappb
 Approach: 
 Cloak
idappb
 Toggle Cloak
iddiagb
  Diagrammatic Representation 
 Cloak
iddiagb
We again sketch the situation and construct a coordinate system.
Image Added
 Cloak
diagb
diagb
 Toggle Cloak
idmathb
  Mathematical Representation 
 Cloak
idmathb
We now write the equations of Newton's 2nd Law for the center of mass and of torque balance about the center of mass for the bicycle. 
 Note
The bicycle's center of mass is accelerating linearly in the negative x direction, but the bicycle is not rotating about its center of mass. Thus the torques about the center of mass must balance.
 Warning
Because the bicycle is accelerating, the only point moving with the bike that can be chosen as an axis of rotation is the center of mass. Try repeating the problem with another axis (such as the point of contact of the front wheel with the ground) and you should find that the torques do not balance).
 Latex
\begin{large} \[ \sum F_{x} = - F_{f,{\rm front}} - F_{f,{\rm rear}} = ma_{x} \]
\[\sum F_{y} = - mg + N_{\rm front} + N_{\rm rear} = 0 \]
\[\sum \tau = N_{\rm front} L_{\rm front} - N_{\rm rear} L_{\rm rear} - F_{f,{\rm front}} h - F_{f,{\rm rear}}h  = 0\]\end{large}
{latex}


We 
 
 have 
 
 only 
 
 three 
 
 equations 
 
 and 
 
 four 
 
 unknowns, 
 
 but 
 
 because 
 
 the 
 
 friction 
 
 forces 
 
 have 
 
 the 
 
 same 
 [
moment 
 
 arm
] 
 about 
 
 the 
 
 center 
 
 of 
 
 mass, 
 
 we 
 
 can 
 
 substitute 
 
 for 
 
 their 
 
 sum. 
  
 Thus, 
 
 using 
 
 torque 
 
 balance, 
 
 we 
 
 can 
 
 find:


{
 Latex
}
\begin{large}\[ N_{\rm front} = \frac{N_{\rm rear}L_{\rm rear} - ma_{x}h}{L_{\rm front}}\]\end{large}
{latex}


We 
 
 can 
 
 then 
 
 substitute 
 
 for 
 _
N
_~rear~ 
rear using 
 
 Newton's 
 
 2nd 
 
 Law 
 
 for 
 
 the 
 _
y
_ 
 direction:


{
 Latex
}
\begin{large}\[ N_{\rm front} = \frac{mg L_{\rm rear} - ma_{x}h}{L_{\rm rear}+L_{\rm front}} = \mbox{718 N}\]
\end{large}
{latex}

{note}Note that _a_~x~ is negative in our coordinate system.{note}

Which means that:

{latex}
 Note
Note that ax is negative in our coordinate system.
Which means that:
 Latex
\begin{large}\[ N_{\rm rear} = mg-N_{\rm front} = \frac{mg L_{\rm front} + ma_{x}h}{L_{\rm rear}+L_{\rm front}} =\mbox{213 N} \]\end{large}
{latex}

{cloak:mathb}
{cloak:appb}
{card}
{card:label=Part C}

h2. Part C

What is the largest coefficient of kinetic friction allowed between the front tire and the ground if the rear tire is to remain on the ground during a skid?

h4. Solution

{toggle-cloak:id=sysc} *System:* {cloak:id=sysc} Cyclist plus bicycle are treated as a single [rigid body].{cloak}

{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc}External forces from the earth (gravity) and the ground (normal force and friction).  {cloak}

{toggle-cloak:id=modc} *Model:*  {cloak:id=modc}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appc} *Approach:* 

{cloak:id=appc}

 The problem is the same as Part B, except that we take the limit as _N_~rear~ approaches zero.  This gives the equations:

{latex}\begin{large} \[ \sum F_{x} = - \mu_{k}N_{\rm front} = ma_{x} \]
\[\sum F_{y} = - mg + N_{\rm front} = 0 \]
\[\sum \tau = N_{\rm front} L_{\rm front} - \mu_{k}N_{\rm front} h  = 0\]\end{large}{latex}

The equation of torque balance is the key here.  If the torques are to sum to zero, we see that we must have:

{latex}\begin{large} \[ \mu_{k} = \frac{L_{\rm front}}{h} = 0.88 \]\end{large}{latex}

This is an upper limit, since a larger force of friction will result in a negative torque, tending to raise the rear wheel.  A lower force of friction will create a positive torque, tending to press the rear wheel against the ground (we should not have neglected the rear normal force if &mu;~k~ < 0.88). 

{tip}A coefficient of friction of 0.88 is high for kinetic friction between tire and road, but reasonable for static friction (realized if the front tire is rolling without slipping).  For this reason, braking hard with the front wheel can be dangerous.{tip}

{cloak}
{card}
{deck}




 Cloak
mathb
mathb
 Cloak
appb
appb
 Card
labelPart C
Part C
What is the largest coefficient of kinetic friction allowed between the front tire and the ground if the rear tire is to remain on the ground during a skid?
Solution
 Toggle Cloak
idsysc
 System: 
 Cloak
idsysc
 Cyclist plus bicycle are treated as a single .
 Toggle Cloak
idintc
 Interactions: 
 Cloak
idintc
External forces from the earth (gravity) and the ground (normal force and friction). 
 Toggle Cloak
idmodc
 Model: 
 Cloak
idmodc
 and .
 Toggle Cloak
idappc
 Approach: 
 Cloak
idappc
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