A yo-yo has a total mass of 250g, distributed among its two disks and axle where

Question

A yo-yo has a total mass of 250g, distributed among its two disks and axle where md=100g and ma=50g. The radii of the disks and axle are rd=3.00cm and ra=1.00cm and all can be modeled as solid cylinders. Initially, a string 3.00ft long with negligible mass is wound about the axle. Holding the end of the string, you release the yo-yo and watch it fall. Air resistance is minimal enough to neglect.

A) As it falls, what is its acceleration?

B) How much time does it take for it to reach the bottom?

C) What is its kinetic energy at the bottom of its motion? is this equivalent to its initial potential energy? why or why not?

D) What is its angular momentum at the bottom of its motion?

E) Assuming that the string is tightly-secured about the axle, such that no slipping occurs, why must the yo-yo begin rising after it has fallen?

Explanation / Answer

Moment of inertia of system = (MaRa^2 + 2MdRd^2)/2 = 9.25 e-5

Since static friction does no work, power equals rate of change of KE

(Ma+Md)g v = (Ma+Md + I/ra^2) x accn x v

accn = 0.213 m/s^2

time taken to travel 3 ft = 0.9144m = sqrt(2x0.9144/0.213) = 2.93 sec

kinetic energy at bottom = initial potential energy because static friction did not do any work = (Ma+Md)g x 0.9144 = 2.24 J

Angular momentum = (Ma+Md + I/ra^2)x ra x v = 7.33 e-3 kgm^2/sec

At the bottom point, the sudden jerk makes the linear and angular momentum 0. However, the tension in the string persists and is equal to the friction force. This force pulls up the yoyo backwards although it does not rise as much due to the loss of energy in the collission.

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