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/s2. the child is standing 4.65 m from the center of the merry-go-round. what angle does the acceleration of the child make with the tangential direction

Answer 1

w = instantaneous angular speed = 1.25 rad/s

r = radius = 4.65 m

α = angular acceleration = 0.745 rad/s²

centripetal acceleration is given as

$a_{c}$ = rw²

$a_{c}$ = (4.65 ) (1.25)² = 7.27 m/s²

tangential acceleration is given as

$a_{t}$ = rα

$a_{t}$ =4.65 x 0.745 = 3.46 m/s²

angle is given as

$\theta = tan^{-1}(\frac{a_{c}}{a_{t}})$

$\theta = tan^{-1}(\frac{7.27}{3.46})$

$\theta$ = 64.5 deg