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 is the magnitude of the acceleration of the child?

Answer 1

The angular speed, ω = 1.25 rad/sThe angular acceleration, α = 0.745 rad/s²The distance, r = 4.65 m------------------------------------------------------------------------------------------------

The relation between the linear velocity and angular velocity is

v = rω

The radial acceleration of the child is

a = v² /r

= r²ω²/r

= rω²

= (4.65 m)(1.25 rad/s)²

= 7.265625 m/s²

The tangential acceleration of the child is

a_t = αr

= (0.745 rad/s²)(4.65 m)

= 3.46425 m/s²

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The magnitude of the acceleration of the child is

a = √a²+a_t²

= √(7.265625 m/s²)²+(3.46425m/s²)²

= 8.049 m/s²

= 8.05 m/s²

Answer 2

There are two accelerations that you have to add asvectors:

radially:    a = v² /r      you need the child's speed whichis

     v = ωr = 1.25 *4.65 = 5.8125   m/s

    so radial acc     a = 5.8125² / 4.65 = 7.266m/s²

and tangentially...     a = αr = 0.745 * 4.65 = 3.464m/s²

Since these are perpendicular to each other, we get the totalacc by adding these like they are legs of a right triangle, usingthe pythagorean theorem

mag of total acc =   (7.266² + 3.464²)^1/2 =    8.05 m/s²