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Excerpt
hiddentrue

Determine the final speed of a ball that is initially sliding without rotation.

Image Modified

Photo courtesy Wikimedia Commons, by user Fir0002

...

Deck of Cards
idbigdeck
h4.

Method

1:

Angular

Momentum

{

Card
labelMethod 1: Angular Momentum
Wiki Markup
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:

id

=

sys1

} *

System:
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idsys1

The ball as a .

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idint1
Interactions:
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idint1

External forces of gravity, normal force and kinetic friction.

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idmod1
Model:
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idmod1

.

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idapp1
Approach:

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idapp1

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iddiag1
Diagrammatic Representation

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iddiag1

We begin with a free body diagram. Since this problem will deal with rotation, we must also select an axis of rotation.

Image Added

In this case, the most direct (but probably not the most obvious) method of solution is to consider the angular momentum about an axis fixed at some point on the alley surface. For simplicity, we select the point at which the ball is released. By taking a point on the alley surface, we have guaranteed that there is zero net torque acting on the system (the ball) about this axis. Gravity and the normal force each create a nonzero torque, but they balance each other perfectly because they share the same moment arm and they have the same magnitude. The friction force produces no torque because it has zero lever arm.

Image Added

Note

If it is not apparent to you that the force of gravity and the normal force acting on the ball have equal magnitudes, you should write Newton's 2nd Law for the y-direction.

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diag1
diag1

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idmath1
Mathematical Representation

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idmath1

The angular momentum about a single axis of the ball about the axis we have chosen can be broken into two parts by considering the contribution of the translation of the ball's center of mass and the rotation of the ball about its center of mass.

Latex
*  {cloak:id=sys1}The ball as a [rigid body].{cloak}

{toggle-cloak:id=int1} *Interactions:* {cloak:id=int1}External forces of gravity, normal force and kinetic friction.{cloak}

{toggle-cloak:id=mod1} *Model:*  {cloak:id=mod1}[Angular Momentum and External Torque about a Single Axis].{cloak}

{toggle-cloak:id=app1} *Approach:*  
{cloak:id=app1}

{toggle-cloak:id=diag1} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diag1}
We begin with a free body diagram.  Since this problem will deal with rotation, we must also select an axis of rotation.

!bowlballfbd.jpg!

In this case, the most direct (but probably not the most obvious) method of solution is to consider the angular momentum about an axis fixed at some point on the alley surface.  For simplicity, we select the point at which the ball is released.  By taking a point on the alley surface, we have guaranteed that there is zero net torque acting on the system (the ball) about this axis.  Gravity and the normal force each create a nonzero torque, but they balance each other perfectly because they share the same [moment arm] and they have the same magnitude.  The friction force produces no torque because it has zero lever arm.

!bowlballleverarm.jpg!

{note}If it is not apparent to you that the force of gravity and the normal force acting on the ball have equal magnitudes, you should write Newton's 2nd Law for the y-direction.{note}

{cloak:diag1}

{toggle-cloak:id=math1} {color:red} *Mathematical Representation* {color}

{cloak:id=math1}
The [angular momentum about a single axis] of the ball about the axis we have chosen can be broken into two parts by considering the contribution of the translation of the ball's center of mass and the rotation of the ball about its center of mass.

{latex}\begin{large}\[ L = L_{\rm trans} + L_{\rm rot} = mRv + I\omega \] \end{large}{latex}

Since

there

is

zero

net

torque

acting

on

the

ball,

the

angular

momentum

is

conserved

and

we

can

write:

{
Latex
}\begin{large}\[ mRv_{f} + I\omega_{f} = mRv_{i}+I\omega_{i} \] \end{large}{latex}

We

know

that ω~i~ =

that ωi = 0,

and

we

also

know

that

the

ball

ends

up

rolling

without

slipping.

Thus,

we

know

that:

{
Latex
}\begin{large}\[ \omega_{f}R = v_{f}\]\end{large}{latex}

Thus,

we

have:

{
Latex
}\begin{large}\[ v_{f} = \frac{v_{i}}{1+\frac{\displaystyle I}{\displaystyle mR^{2}}} = \frac{v_{i}}{1+\frac{\displaystyle 2}{\displaystyle 5}} = \frac{5}{7} v_{i} = \mbox{2.0 m/s} \]\end{large}{latex}

{cloak:math1}
{cloak:app1}

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Card
labelMethod 2: Dynamics

Method 2: Dynamics

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idsys2
System:
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idsys2

The ball as a .

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idint2
Interactions:
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idint2

External forces of gravity, normal force and kinetic friction.

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idmod2
Model:
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idmod2

Single-Axis Rotation of a Rigid Body, and One-Dimensional Motion with Constant Acceleration.

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idapp2
Approach:

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idapp2
Card
labelMethod 3: Energy

Method 3: Energy

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idsys3
System:
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idsys3

The ball as a .

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idint3
Interactions:
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idint3

External forces of gravity, normal force and kinetic friction. Gravity is conservative while the normal force and friction are non-conservative.

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idmod3
Model:
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idmod3

Mechanical Energy, External Work, and Internal Non-Conservative Work, One-Dimensional Motion with Constant Acceleration and .

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idapp3
Approach:

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idapp3

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