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Deck of Cards

id	bigdeck

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

label	Part 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

id	moda

and .

Toggle Cloak

id	appa

Approach:

Cloak

id	appa

Toggle Cloak

id	diaga

Diagrammatic Representation

Cloak

id	diaga

We first sketch the situation and construct a coordinate system.

Image Added

Cloak

	diaga
	diaga

Toggle Cloak

id	matha

Mathematical Representation

Cloak

id	matha

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

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

Card

label	Part B

Wiki Markup

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

and .

Toggle Cloak

id	appb

Approach:

Cloak

id	appb

Toggle Cloak

id	diagb

Diagrammatic Representation

Cloak

id	diagb

We again sketch the situation and construct a coordinate system.

Image Added

Cloak

	diagb
	diagb

Toggle Cloak

id	mathb

Mathematical Representation

Cloak

id	mathb

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 a_x 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

label	Part 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

id	sysc

System:

Cloak

id	sysc

Cyclist plus bicycle are treated as a single .

Toggle Cloak

id	intc

Interactions:

Cloak

id	intc

External forces from the earth (gravity) and the ground (normal force and friction).

Toggle Cloak

id	modc

Model:

Cloak

id	modc

and .

Toggle Cloak

id	appc

Approach:

Cloak

id	appc

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Key

Part

A

Solution

Part

B

Solution

Part C

Solution