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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

Respuesta :

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

[tex] a_{c} [/tex] = rw²

[tex] a_{c} [/tex] = (4.65 ) (1.25)² = 7.27 m/s²

tangential acceleration is given as

[tex] a_{t} [/tex] = rα

[tex] a_{t} [/tex] =4.65 x 0.745 = 3.46 m/s²

angle is given as

[tex] \theta = tan^{-1}(\frac{a_{c}}{a_{t}}) [/tex]

[tex] \theta = tan^{-1}(\frac{7.27}{3.46}) [/tex]

[tex] \theta [/tex] = 64.5 deg

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