Question

Suppose a yo-yo has a center shaft that has a 2.70 cm radius and that its string is being pulled. If the string is stationary and the yo-yo accelerates away from it at a rate of 1.65 m/s2, what is the angular acceleration of the yo-yo

Answer 1

Answer:

$\alpha= \frac{1.65 m/s^2}{0.027 m}= 61.11 rad/s^2$

Explanation:

For this case we can use the following formula for the angular velocity:

$w = w_o + \alpha t$

where $w$ represent the final angular velocity , $w_o$ the initial angular velocity , t the time and $\alpha$ the angular acceleration.

And for the linear acceleration we have this formula:

$a = r \alpha$

Where a represent the linear acceleration and $\alpha$ the angular acceleration.

For this case the linear acceeleration is given $a_c = 1.65 m/s^2$

And the radius of the yo-yo is also given $r= 2.7 cm= 0.027 m$

So then we can use the following formula:

$a = r \alpha$

If we replace the values we got:

$1.65m/s^2 = 0.027 m (\alpha)$

And solving for $\alpha$ we got:

$\alpha= \frac{1.65 m/s^2}{0.027 m}= 61.11 rad/s^2$