Answer:
(a) [tex]v=31.16\frac{m}{s}[/tex]
(b) [tex]a_r=29.14g[/tex]
Explanation:
(a) The linear speed in a circular motion is defined as the distance traveled in one revolution divided into the time taken to complete one revolution. So, we have:
[tex]v=\frac{2\pi r}{T}[/tex]
Here r is the radius of the circular motion, that is, the distance from the central shaft to the blade tip. As can be seen this time is the period, which is defined as:
[tex]T=\frac{2\pi}{\omega}[/tex]
[tex]\omega[/tex] is the angular speed. Replacing this in the linear speed equation:
[tex]v=\frac{2\pi r}{\frac{2\pi}{\omega}}\\v=\omega r\\v=550\frac{rev}{min}(3.40m)\\v=1870\frac{m}{min}*\frac{1min}{60s}\\v=31.16\frac{m}{s}[/tex]
(b) The radial acceleration is given by:
[tex]a_r=\frac{v^2}{r}\\a_r=\frac{(31.16\frac{m}{s^2})^2}{3.40m}\\a_r=285.57\frac{m}{s^2}\\\\\frac{a_r}{g}=\frac{285.57\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\frac{a_r}{g}=29.14\\a_r=29.14g[/tex]
