Answer:
Acceleration, [tex]a=15.68\ m/s^2[/tex]
Explanation:
It is given that, angle rotated by blade is, [tex]\theta=90^{\circ}=\dfrac{\pi}{2}[/tex] (because [tex]180^{\circ}=\pi\ radians[/tex])
Also, time taken , t =0.28 s.
It is also given that , the radius of arc in which it moves is, r = 0.5 m.
Time taken to move in a circle, [tex]T=0.28\times 4=1.12\ s[/tex]
Now , we need to find centripetal acceleration. Its formula is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
Since, [tex]v=\dfrac{2\pi r}{T}[/tex]
[tex]v=\dfrac{2\pi \times 0.5}{1.12}[/tex]
v = 2.80 m/s
So, [tex]a=\dfrac{(2.8)^2}{0.5}[/tex]
[tex]a=15.68\ m/s^2[/tex]
So, the magnitude of the centripetal acceleration of the tip of the blade is [tex]15.68\ m/s^2[/tex]. Hence, this is the required solution.
