Question

The blade of a windshield wiper moves through an angle of 90.0° in 0.397 s. The tip of the blade moves on the arc of a circle that has a radius of 0.376 m. What is the magnitude of the centripetal acceleration of the tip of the blade?

Answer 1

Answer:

$a=5.88\ m/s^2$

Explanation:

Given that,

The blade of a windshield wiper moves through an angle of 90.0° in 0.397 s

The radius of the circlular path is 0.376 m

We need to find the magnitude of the centripetal acceleration of the tip of the blade. Let it is a. The formula of the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

v is velocity, $v=\dfrac{2\pi r}{t\\}$

It moves through an angle of 90 degrees, it means $v=\dfrac{\dfrac{1}{4}\times 2\pi r}{t\\}$

So,

$a=\dfrac{(\dfrac{\dfrac{1}{4}\times 2\pi r}{t\\})^2}{r}\\\\=\dfrac{\pi^2 r}{4t^2}\\\\=\dfrac{\pi^2 \times 0.376}{4(0.397 )^2}\\\\=5.88\ m/s^2$

So, the magnitude of the centripetal acceleration of the tip of the blade is $5.88\ m/s^2$.