Question

The aorta is a major artery, rising upward from the left ventricle of the heart and curving down to carry blood to the abdomen and lower half of the body. The curved artery can be approximated as a semicircular arch whose diameter is 5.8 cm. If blood flows through the aortic arch at a speed of 0.34 m/s, what is the magnitude (in m/s2) of the blood's centripetal acceleration?

Answer 1

The centripetal acceleration of the blood is
`3.98\ m/s^(2)`

Solution:

As per the question:

Diameter of the semi-circular arc of the curved artery, d = 5.8 cm = 0.058 m

The speed with which the blood flows, v = 0.34 m/s

Now, the centripetal acceleration of the blood through the curved artery is given by:

By Newton's second law:

F = ma

Also, Centripetal force:

`F_(C) = (mv^(2))/(r)`

Now,

`ma = (mv^(2))/(r)`

Thus the centripetal acceleration is given by:

`a = (v^(2))/(r)`

`a = (v^(2))/((d)/(2))`

`a = ((0.34)^(2))/((0.058)/(2)) = 3.98 m/s^(2)`