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

`\alpha= (1.65 m/s^2)/(0.027 m)= 61.11 rad/s^2`

Step-by-step 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= (1.65 m/s^2)/(0.027 m)= 61.11 rad/s^2`