Question

Suppose a yo-yo has a center shaft that has a 0.230 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 1.80 m/s2, what is the angular acceleration of the yo-yo in rad/s2? rad/s2.
(b) What is the angular velocity in rad/s after 0.750 s if it starts from rest? rad/s.
(c) The outside radius of the yo-yo is 3.10 cm.What is the tangential acceleration in m/s2 of a point on its edge? m/s2

Answer 1

Part a)

`\alpha = 782.6 rad/s^2`

Part B)

`\omega = 587 rad/s`

Part c)

`a_t = 24.3 m/s^2`

Step-by-step explanation:

Part a)

As we know that

`a = R \alpha`

so we will have

`a = 1.80 m/s^2`

`R = 0.230 cm`

`\alpha = (a)/(R)`

`\alpha = (1.80)/(0.230 * 10^(-2))`

`\alpha = 782.6 rad/s^2`

Part B)

Angular speed of the yo-yo

`\omega = \alpha t`

so we have

`\omega = 782.6 * 0.750`

`\omega = 587 rad/s`

Part c)

Tangential acceleration is given as

`a_t = R \alpha`

`a_t = (3.10 * 10^(-2))(782.6)`

`a_t = 24.3 m/s^2`