Question

The blade of a windshield wiper moves through an angle of 180.0 degrees in 0.500 s. The tip of the blade moves on the arc of a circle of radius of 0.520 m. What is the magnitude of the centripetal acceleration of the tip of the blade

Answer 1

Answer:

Centripetal acceleration will be equal to $10.66m/sec^2$

Explanation:

We have given time taken to cover 180° is 0.5 sec

So time taken by 360° is equal to = 2×0.5 = 1 sec

Radius of the circle r = 0.520 m

So distance $d=2\pi r=2\times3.14\times 0.520=3.2656m$

So velocity $v=\frac{distance}{time}=\frac{.2656}{1}=3.2656m/sec$

We have to find the centripetal acceleration

Centripetal acceleration will be equal to $a_c=\frac{v^2}{r}=\frac{3.2656^2}{1}=10.66m/sec^2$

So centripetal acceleration will be equal to $10.66m/sec^2$