Suppose a yo-yo has a center shaft that has a 0.200 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 1.70 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 3.30 cm. What is the tangential acceleration in m/s2 of a point on its edge?

Respuesta :

Answer:

a) [tex]\alpha =850rad/s^2[/tex]

b) [tex]\omega = 637.5rad/s[/tex]

c) [tex]a =29.75 m/s^2[/tex]

Explanation:

The angular acceleration is given by:

[tex]\alpha = a/r[/tex]

[tex]\alpha = 1.7 / 0.002[/tex]

[tex]\alpha = 850rad/s^2[/tex]

Angular velocity is calculated by kinematics:

[tex]\ omega = \alpha*t[/tex]

[tex]\ omega = 850*0.75[/tex]

[tex]\omega = 637.5rad/s[/tex]

Accelerarion on the opposite edge of the string is:

[tex]a = \alpha*(r+R)[/tex]

[tex]a = 850 * ( 0.002+0.033)[/tex]

[tex]a = 29.75m/s^2[/tex]

ACCESS MORE
EDU ACCESS