183_notes:examples:a_yo-yo [Projects & Practices in Physics]

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===== Example: A Yo-yo  =====

You're playing with a yo-yo of mass m on a low-mass string (See Diagram in Representations). You pull up on the string with a force of magnitude F, and your hand moves up a distance d. During this time the mass falls a distance h (and some of the string reels off the yo-yo's axle). 

(a) What is the change in translational kinetic energy of the yo-yo?

(b) What is the change in the rotational kinetic energy of the yo-yo, which spins faster?

=== Facts ===

a:

Initial State: Point particle with initial translational kinetic energy

Final State: Point particle with final translational kinetic energy


b: 

Initial State: Initial rotational and translational kinetic energy

Final State: Final rotational and translational kinetic energy


=== Assumptions and Approximations ===

You are able to maintain constant force when pulling up on yo-yo

Assume no slipping of string around the axle. Spindle turns the same amount as string that has unravelled

No wobble included

String has no mass


=== Lacking ===

Change in translational kinetic energy of the yo-yo

Change in the rotational kinetic energy of the yo-yo

=== Representations ===

a:

Point Particle System

System: Point particle of mass $m$

Surroundings: Earth and hand

{{course_planning:course_notes:mi3e_09-034.jpg?100|}}

b: 

Real system

System: Mass and string

Surroundings: Earth and hand

{{course_planning:course_notes:mi3e_09-035.jpg?100|}}

$\Delta K_{trans}$ = $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$

$\Delta E_{sys}$ = $W_{surr}$

=== Solution ===

a: 

From the Energy Principle ( when dealing with a point particle it only has $K_{trans}$):

$\Delta K_{trans}$ = $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$

Substituting in for the forces acting on the yo-yo for $F_{net}$ and the change in position in the y direction for the centre of mass for $d\vec{r}_{cm}$ we get:

$\Delta K_{trans} = (F - mg)\Delta y_{CM}$

As indicated in diagram in the b section of the representation:

$\Delta y_{CM} = -h$ 

Substitute in $-h$ for $y_{CM}$

$\Delta K_{trans} = (F - mg)(-h)$

Multiply across by a minus and you get an equation for $\Delta K_{trans}$ that looks like:  

$\Delta K_{trans} = (mg - F)h$


b: 

From the energy principle we know:

$\Delta E_{sys}$ = $W_{surr}$

In this case we know that the change in energy in the system is due to the work done by the hand and the work done by the Earth.

$\Delta E_{sys} = W_{hand} + W_{Earth}$

Because we are dealing with the real system in this scenario the change in energy is equal to the change in translational kinetic energy + the change in rotational kinetic energy.

$\Delta K_{trans} + \Delta K_{rot} = W_{hand} + W_{Earth}$

Substitute in the work represented by force by distance for both the hand and the Earth.

$\Delta K_{trans} + \Delta K_{rot} = Fd + (-mg)(-h)$

From part (a) of the problem we can substitute in $(mg - F)h$ for $\Delta K_{trans}$ as the translational kinetic energy will be the same.

$\Delta K_{trans} = (mg - F)h$ 

Substituting this into our equation leaves us with:

$(mg - F)h + \Delta K_{rot} = Fd + mgh$

Solve for change in rotational kinetic energy:

$\Delta K_{rot} = F(d + h)$