As a yo-yo master, you have a selection of yo-yo\'s to choose from. For this fea

Question

As a yo-yo master, you have a selection of yo-yo's to choose from. For this feat of amazement, you've chosen a different yo-yo, with mass 51.0 g, radius 1.79 cm, with an axle radius of 0.46 cm, but otherwise can be modeled exactly like the previous problem. Once again, you exhibit your skill by allowing the yo-yo to slowly unravel, but this time, you pull the end of the string upward with a constant force, such that the yo-yo spins, but is suspended in place for a short time, defying gravity itself! What is the angular acceleration of the yo-yo during this time? Hint: Think carefully about what your free-body diagrams are telling you!

Explanation / Answer

STEP ONE: let us derive all the required things for YOYO

Let:

'm' be the mass of each disk,

'R' be the radius of each disk,

'b' be the radius of the light axle,

'g' be the acceleration due to gravity,

'T' be the tension in the string,

'I' be the moment of inertia of the yo-yo about its centre,

'alpha' be the angular acceleration.

The equation of motion for the centre of mass is:

2mg - T = 2ma ...(1)

The equation for the rotational motion is:

I alpha = bT ...(2)

In addition:

a = b alpha ...(3)

I = mR^2 ...(4)

Eliminating 'T' from (1) and (2):

I alpha = 2mb(g - a) ...(5)

Eliminating alpha from (3) and (5):

Ia = 2mb^2 (g - a)

(I + 2mb^2)a = 2mb^2 g

a = 2mb^2 g / (I + 2mb^2).

Now, given;

m = 51.0 g = 0.051 kg;

R = 1.79 cm = 0.0179 m;

b = 0.46 cm = 0.0046 m;

I = mR2 = 1.63*10-5;

a =  2mb2 g / (I + 2mb2) = [2*0.051*0.00462*9.81]/[1.63*10-5 + (2*0.051*0.00462)]

a = 1.1471 m/s2

from eq(3), angular acceleration, alpha = a/b = 1.1471 m/s2 / 0.0046 m = 249.3646

the angular acceleration of the yo-yo, alpha = 39.69 rad/s2

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