Question

Cheryl is riding on the edge of a merry-go-round, 2m from the center, which is rotating with an increasing angular speed. Cheryl’s tangential acceleration is 3.0m/s2. At the instant that Cheryl’s linear speed is 4.0m/s, what is Cheryl’s total acceleration?

Answer 1

In circular motion we know that there is two type of acceleration that Cheryl experience

1. Tangential acceleration

2. Centripetal acceleration

here given that

$a_t = 3 m/s^2$

for centripetal acceleration we know that

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{2} = 8 m/s^2$

now we know that both centripetal acceleration and tangential acceleration is perpendicular to each other

so total acceleration is vector sum of both and given as

$a_{net} = \sqrt{8^2 + 3^2} = 8.54 m/s^2$

so total acceleration is 8.54 m/s^2