Question

what is the magnitude of the total acceleration of the tip of a 1.20 - m (radius) airplane propeller that is moving with an angular speed of 6.00 rad/s and undergoing an angular acceleration of 14.0 rad/s²?
16.8 m/s²
43.2 m/s²
46.4 m/s²
138 m/s²

Explanation:

Step1: Calculate centripetal acceleration

Centripetal acceleration formula: \(a_{c}=\omega^{2}r\). Given \(\omega = 6.00\ rad/s\) and \(r=1.20\ m\), so \(a_{c}=(6.00)^{2}\times1.20 = 36\times1.20=43.2\ m/s^{2}\)

Step2: Calculate tangential acceleration

Tangential acceleration formula: \(a_{t}=\alpha r\). Given \(\alpha = 14.0\ rad/s^{2}\) and \(r = 1.20\ m\), so \(a_{t}=14.0\times1.20 = 16.8\ m/s^{2}\)

Step3: Calculate total acceleration

Total acceleration formula: \(a=\sqrt{a_{c}^{2}+a_{t}^{2}}\). Substitute \(a_{c}=43.2\ m/s^{2}\) and \(a_{t}=16.8\ m/s^{2}\), we get \(a=\sqrt{43.2^{2}+16.8^{2}}=\sqrt{1866.24 + 282.24}=\sqrt{2148.48}\approx46.4\ m/s^{2}\)

Answer:

\(46.4\ m/s^{2}\) (the third option)