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Deck of Cards
idbigdeck
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 {

Solution

Card
labelPart A
Wiki Markup
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
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
*  {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}\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}

Force balance in the y direction then gives:

Latex
{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}
Cloak
matha
matha

Cloak
appa
appa

unmigrated-wiki-markup
Card
labelPart 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 {

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
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
*  {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}\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}
Note

Note that ax is negative in our coordinate system.

Which means that:

Latex
{latex}

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

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}
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

...