Question

suppose a yo - yo has a center shaft that has a 0.290 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 1.40 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 3.70 cm. what is the tangential acceleration in m/s² of a point on its edge?
m/s²

Explanation:

Step1: Convert radius to meters

Given \(r = 0.290\ cm=0.290\times10^{- 2}\ m\). The formula relating linear acceleration \(a\) and angular acceleration \(\alpha\) is \(a = r\alpha\).

Step2: Solve for angular acceleration (part a)

Rearrange \(a = r\alpha\) to \(\alpha=\frac{a}{r}\). Substitute \(a = 1.40\ m/s^{2}\) and \(r = 0.290\times 10^{-2}\ m\)
\(\alpha=\frac{1.40}{0.290\times10^{-2}}\)
\(\alpha=\frac{1.40}{0.0029}\approx482.76\ rad/s^{2}\)

Step3: Find angular velocity (part b)

The formula for angular velocity \(\omega\) when starting from rest (\(\omega_{0} = 0\)) is \(\omega=\omega_{0}+\alpha t\). Substitute \(\omega_{0} = 0\), \(\alpha\approx482.76\ rad/s^{2}\) and \(t = 0.750\ s\)
\(\omega=(0)+482.76\times0.750 = 362.07\ rad/s\)

Step4: Calculate tangential acceleration (part c)

The outside radius \(R = 3.70\ cm = 3.70\times10^{-2}\ m\). The formula for tangential acceleration \(a_{t}\) is \(a_{t}=R\alpha\). Substitute \(R = 3.70\times10^{-2}\ m\) and \(\alpha\approx482.76\ rad/s^{2}\)
\(a_{t}=3.70\times10^{-2}\times482.76\)
\(a_{t}=17.86\ m/s^{2}\)

Answer:

a. \(483\ rad/s^{2}\)
b. \(362\ rad/s\)
c. \(17.9\ m/s^{2}\)