Answer:
[tex]\alpha= \frac{1.65 m/s^2}{0.027 m}= 61.11 rad/s^2[/tex]
Explanation:
For this case we can use the following formula for the angular velocity:
[tex] w = w_o + \alpha t[/tex]
where [tex] w[/tex] represent the final angular velocity , [tex] w_o[/tex] the initial angular velocity , t the time and [tex]\alpha[/tex] the angular acceleration.
And for the linear acceleration we have this formula:
[tex] a = r \alpha[/tex]
Where a represent the linear acceleration and [tex]\alpha [/tex] the angular acceleration.
For this case the linear acceeleration is given [tex] a_c = 1.65 m/s^2[/tex]
And the radius of the yo-yo is also given [tex] r= 2.7 cm= 0.027 m[/tex]
So then we can use the following formula:
[tex] a = r \alpha[/tex]
If we replace the values we got:
[tex] 1.65m/s^2 = 0.027 m (\alpha)[/tex]
And solving for [tex]\alpha[/tex] we got:
[tex]\alpha= \frac{1.65 m/s^2}{0.027 m}= 61.11 rad/s^2[/tex]
