Question

1. Read Section 11-3: we will consider the motion of a yo-yo moving downward on its string.

(a) In terms of the yo-yo’s mass m, its moment of inertia about the center-of-mass I, and the length of the string l, what is the angular speed ? of a “sleeping” yo-yo that started from rest and just finished rolling all the way down its string? You may assume that kinetic friction between the yo-yo and the string has not yet significantly slowed the yo-yo.

(b) A yo-yo consists of two flat disks of uniform density surrounding a low-mass axle. Each disk has mass m/2, the axle has radius r, and the yo-yo has radius R. What is the angular speed of the yoyo when it is halfway down the string? (Hint: consider Equation 11.2 from your textbook, but think carefully about what radius you should use.)

(c) In 1980, a large yo-yo was released from a crane over San Francisco Bay. The 116-kg yo-yo consisted of two uniform disks of radius 32 cm connected connected by an axle of radius 3.2 cm. What was the magnitude of the yo-yo’s acceleration during (1) its fall? What was the tension in the yo-yo’s cord? Be careful of your signs.

Answer 1

(A) KE_yoyo = m v^2 / 2 + I w^2 /2

and v = w r

KE = m w^2 r^2 / 2 + I w^2 / 2

Applying energy conservation,

PEi + Kei = PEf + KEf

m g L + 0 = 0 + (m r^2 + I) w^2 / 2

w = sqrt(2 m g L / (m r^2 + I )) .....Ans

(B) v = w r and I = 2(m/2)(R^2 /2) = m R^2 / 2

m g (L/2) = m (w r)^2 /2 + (m R^2 / ) w^2 / 2

g L = w^2 r^2 + R^2 w^2

w = sqrt(g L / (r^2 + R^2))

(c) m g - T = m a

and torque = I alpha

(0.032)T = (116 x 0.32^2 / 2)(a/0.032)

T = 11600 a

putting in 1st equation,

116 x 9.8 = (116 + 11600) a

a = 0.097 m/s^2

Tension = 11600a = 1125 N