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Deck of Cards
idbigdeck
h4.

Method

1:

Energy

{

Card
labelMethod 1: Energy
Wiki Markup
Toggle Cloak

:

id

=

sys1

} *

System:

* {cloak:id=sys1}The object as a [point particle] undergoing [pure rotation].{cloak} {toggle-cloak:id=int1} *Interaction:* {cloak:id=int1}External forces from the string (tension,

Cloak
idsys1

The object as a undergoing .

Toggle Cloak
idint1
Interaction:
Cloak
idint1

External forces from the string (tension, non-conservative),

the

earth

(gravity,

conservative)

and

the

table

(normal

force,

non-conservative).

{cloak} {

Toggle Cloak

:

id

=

mod1

} *

Model:
Cloak
idmod1

plus .

Toggle Cloak
idapp1
Approach:

Cloak
idapp1

Toggle Cloak
idpart1
Consider the Energy

Cloak
idpart1

The table is level, so the gravitational potential energy will be constant. We can set it to zero by taking the height of the tabletop to be zero. The normal force is perpendicular to the motion of the object and so does no work. The only force capable of performing work on the object is the tension, which is equal to the force from the person pulling on the string. Thus, the work done by tension will equal the work done by the person. This work can be computed by finding the change in mechanical energy of the object:

Latex
*  {cloak:id=mod1}[Angular Momentum and External Torque about a Single Axis] plus [Mechanical Energy and Non-Conservative Work].{cloak}

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

{cloak:id=app1}

{toggle-cloak:id=part1} {color:red} *Consider the Energy* {color}

{cloak:id=part1}

The table is level, so the [gravitational potential energy|gravitation (universal)] will be constant.  We can set it to zero by taking the height of the tabletop to be zero.   The normal force is perpendicular to the motion of the object and so does no work.  The only force capable of performing work on the object is the tension, which is equal to the force from the person pulling on the string.  Thus, the work done by tension will equal the work done by the person.  This work can be computed by finding the change in mechanical energy of the object:

{latex}\begin{large}\[ W^{NC} + E_{i} = E_{f}\]\end{large}{latex}

or,

since

the

height

of

the

object

is

constant

throughout

the

motion

and

no

spring

is

present:

{
Latex
}\begin{large}\[ W^{NC} = E_{f} - E_{i} = K_{f} - K_{i}\]\end{large}{latex}

Because

the

object

can

be

considered

to

be

moving

in

pure

rotation

about

the

center

of

the

circle,

we

can

compute

its

kinetic

energy

using

the

rotational

formula:

{
Latex
}\begin{large}\[ K = K_{\rm rot} = \frac{1}{2}I\omega^{2}\]\end{large}{latex}

For

a

point

particle,

{
Latex
}\begin{large}\[ I = mr^{2}\]\end{large}{latex}

and

so:

{
Latex
}\begin{large}\[ W = \frac{1}{2}m\left(r_{f}^{2}\omega_{f}^{2} - r_{i}^{2}\omega_{i}^{2}\right)\]\end{large}{latex}

{warning}Note that we have assumed the mass is constant, but we cannot
Warning

Note that we have assumed the mass is constant, but we cannot assume the angular speed is constant.

Cloak
part1
part1

Toggle Cloak
idpart2
Consider the Angular Momentum

Cloak
idpart2

We cannot yet substitute into the equation obtained by considering the object's energy. We do already have the initial and final radius of the motion, but we do not yet have the angular speed. To find it, we consider the torque on the system. We choose the natural location for the axis by locating it at the center of the circle.

Image Added

With this choice, we can see that there is zero net torque on the system, since the moment arm for the tension is zero. With the net torque equal to zero, we are in the special case of constant angular momentum, so:

Latex
 assume the angular speed is constant.{warning}

{cloak:part1}
{toggle-cloak:id=part2} {color:red} *Consider the Angular Momentum* {color}

{cloak:id=part2}
We cannot yet substitute into the equation obtained by considering the object's energy.  We do already have the initial and final radius of the motion, but we do not yet have the angular speed.  To find it, we consider the torque on the system.  We choose the natural location for the axis by locating it at the center of the circle.  

!puckonstringfbd.jpg!

With this choice, we can see that there is zero net torque on the system, since the [moment arm] for the tension is zero.  With the net torque equal to zero, we are in the special case of constant angular momentum, so:

{latex}\begin{large}\[ I_{f}\omega_{f} = I_{i}\omega_{i} \]\end{large}{latex}

By

using

the

formula

for

the

moment

of

inertia

of

a

point

particle,

we

can

show:

{
Latex
}\begin{large}\[ \omega_{f} = \omega_{i}\left(\frac{r_{i}}{r_{f}}\right)^{2}\]\end{large}{latex}

Substituting

into

the

work-energy

theorem

then

gives:

{
Latex
}\begin{large}\[ W = \frac{1}{2} mr_{i}^{2}\omega_{i}^{2}\left(\frac{r_{i}^{2}}{r_{f}^{2}} - 1\right) =\mbox{13.5 J}\]\end{large}{latex}

{cloak:part2}
{cloak:app1}
Cloak
part2
part2

Cloak
app1
app1

Card
labelMethod 2: Dynamics

Method 2: Dynamics

Toggle Cloak
idsys2
System:
Cloak
idsys2

The object as a undergoing .

Toggle Cloak
idint2
Interactions:
Cloak
idint2

External forces from the string (tension), the earth (gravity) and the table (normal force).

Toggle Cloak
idmod2
Model:
Cloak
idmod2

, plus the definition of .

Toggle Cloak
idapp2
Approach:

Cloak
idapp2