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?

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

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

Change in translational kinetic energy of the yo-yo

Change in the rotational kinetic energy of the yo-yo

a:

Point Particle System

System: Point particle of mass $m$

Surroundings: Earth and hand

b:

Real system

System: Mass and string

Surroundings: Earth and hand

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

a:

From the Energy Principle (point particle only has $K_{trans}$):

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

$\Delta y_{CM} = -h (from\; digram)$

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

b:

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

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

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

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

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