Question

A child is riding a merry-go-round, which has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s 2 . The child is standing 4.65 m from the center of the merrygo-round.

What is the magnitude of the acceleration of the child?

Answer 1

8.05 m/s^2

Step-by-step explanation:

Instantaneous angular speed(w) = 1.25 rad/s

Angular acceleration (α) = 0.745 rad/s^2

Distance from the centre to where the child is (r) = 4.65m

Tangential acceleration = αr

= 0.745 * 4.65

= 3.464 m/s^2

Centripetal acceleration = w^2r

= (1.25)^2(4.65)

= 7.266 m/s^2

Total acceleration = √3.464^2 + 7.266^2

= √64.794052

= 8.05m/s^2