Question

Suppose a yo-yo has a center shaft that has a 0.28 cm radius and that its string is being pulled.

a. If the string is stationary and the yo-yo accelerates away from it at a rate of 1.2 m/s2 , what is the angular acceleration of the yo-yo?

b. What is the angular velocity after 0.75 s if it starts from rest?

c. The outside radius of the yo-yo is 3.5 cm. What is the tangential acceleration of a point on its edge?

Answer 1

Solution) r1 = 0.28 cm = 0.28×10^(-2) m

(a) a = 1.2 m/s^2

Angular acceleration = ?

a = (r1)(Angular acceleration)

Angular acceleration = (a)/(r1)

Angular acceleration = (1.2)/(0.28×10^(-2))

Angular acceleration = 428.57 rad/s^2

(b) t = 0.75 s

w = ?

wo = 0 (rest)

w = wo + (Angular acceleration)(t)

w = 0 + (Angular acceleration)(t)

w = (428.57)(0.75)

w = 321.4 rad/s

(c) r2 = 3.5 cm = 3.5×10^(-2) m

Tangential acceleration , a = ?

a = (r2)(Angular acceleration)

a = (3.5×10^(-2))(428.57)

a = 14.99

a = 15 m/s^2