Question

A child riding mary go round complete one revolution every 8.04 s the child standing 4.17 m from the center of marygoround what is the magnitude of centripetal acceleration of child ?

Answer 1

The magnitude of the child's centripetal acceleration is 2.53
`(m)/(s^(2) )`

Step-by-step explanation:

Centripetal acceleration is calculated with the following formula:

`a_(c) =w^(2) *r`

`a_(c)`= Centripetal acceleration (
`(m)/(s^(2) )`)

w= speed angular (
`(rad)/(s)`)

r= radius of the curve(m)

Calculation of the angular velocity of the child

`w=(1 revolution)/(8.04s)` equation (1)

`1revolution=2*\pi`, then, in the equation(1):

`w=(2*\pi )/(8.04)`

`w=0.78(rad)/(s)`

Calculation of the child's centripetal acceleration

We replaced
`w=0.78(rad)/(s)`and
`r=4.17m`in the equation (1)

`a_(c) =0.78^(2) *4.17`

`a_(c) =0.6084*4.17`

`a_(c) =2.53(rad)/(s^(2) )`

Answer: The magnitude of the child's centripetal acceleration is
`2.53(m)/(s^(2) )`