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

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

Step-by-step 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*3.14* 0.520=3.2656m`

So velocity
`v=(distance)/(time)=(.2656)/(1)=3.2656m/sec`

We have to find the centripetal acceleration

Centripetal acceleration will be equal to
`a_c=(v^2)/(r)=(3.2656^2)/(1)=10.66m/sec^2`

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