9. [1pt] Three yo-yos (labelled 1, 2, and 3 in the figure) have the same outer d

Question

9. [1pt]
Three yo-yos (labelled 1, 2, and 3 in the figure) have the same outer diameter R, but the string is wrapped around axles of various diameters, as shown. The yo-yos are released from rest and from the same height, at the same time. Ignore the mass of the axle. Think first before answering true, false, or cannot tell to the statements below. E.g., if the first statement is true, the second cannot be determined, and the rest are false, enter TCFFFF.

i) As yo-yo 1 falls, the velocity of its center of mass is related to its angular velocity by vcm = r.

ii) The yo-yos spin counter-clockwise about their centers as they fall.

iii) After release, the tension in the string of yo-yo 1 is greater than the tension in the string of yo-yo 2.

iv) As the yo-yos fall a distance h, gravitational potential energy is converted only to rotational kinetic energy.

v) The angular velocity of yo-yo 3 is the largest, after all yo-yos have fallen a distance h.

vi) Yo-yo 2 is the first to fall a distance h.

For the next 3 questions, consider a yo-yo with R = 4.4 cm, r = 0.3 cm, and mass m = 65 g.

10. [1pt]
What is the speed of the yo-yo after it has fallen a distance h = 1.31 m?

11. [1pt]

What is the tension in the string while the yo-yo falls?

12. [1pt]
What is the magnitude of the acceleration of the center of mass as the yo-yo falls?

Explanation / Answer

9.(1)

As yo-yo 1 falls, the velocity of its center of mass,

vcm = wr (TRUE)

(2)

yo-yos spin clockwise about their centers as they fall (FALSE)

(3)

FALSE

(4)

As the yo-yos fall a distance h, gravitational potential energy is converted only to rotational kinetic energy.

TRUE

(5)

linear velocity will be smallest for 3 hence angular velocity will laregst.

TRUE

(6)

yo-yo 1 will fall first. (FALSE).

answer is TFFTTF.

(10.)

Here,

mg - T = ma

T*r = l*alpha = I*(a/r)

so, T = (I / r^2) * a

mg = ma + T

so, a = mg / (m + (I / r^2))

(l = mR^2 / 2)

a = 0.065*9.8 / (0.065 + (0.065*0.044^2 / 2)/0.003^2)

a = 0.0902 m/s^2

From equation of motion,

v^2 = u^2 + 2ah

v^2 = 0 + 2*0.0902*1.31

v = 0.486 m/s

(11.)

T = (I / r^2) * a

T =  ((0.065*0.044^2 / 2)/0.003^2) * 0.0902

T = 0.63 N

(12.)

magnitude of the acceleration of the center of mass,

a = 0.0902 m/s^2

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