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
Surroundings: Earth and hand
b:
Real system
System: Mass and string
Surroundings: Earth and hand
=
=
a:
From the Energy Principle ( when dealing with a point particle it only has ):
=
Substituting in for the forces acting on the yo-yo for and the change in position in the y direction for the centre of mass for we get:
As indicated in diagram in the b section of the representation:
Substitute in for
Multiply across by a minus and you get an equation for that looks like:
b:
From the energy principle we know:
=
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.
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.
Substitute in the work represented by force by distance for both the hand and the Earth.
From part (a) of the problem we can substitute in for as the translational kinetic energy will be the same.
Substituting this into our equation leaves us with:
Solve for change in rotational kinetic energy: