contestada

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 ?

Respuesta :

Answer:

The magnitude of the child's centripetal acceleration is 2.53 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Centripetal acceleration is calculated with the following formula:

[tex]a_{c} =w^{2} *r[/tex]

[tex]a_{c}[/tex]= Centripetal acceleration ([tex]\frac{m}{s^{2} }[/tex])

w= speed angular ([tex]\frac{rad}{s}[/tex])

r= radius of the curve(m)

Calculation of the angular velocity of the child

[tex]w=\frac{1 revolution}{8.04s}[/tex] equation (1)

[tex]1revolution=2*\pi[/tex], then, in the equation(1):

[tex]w=\frac{2*\pi }{8.04}[/tex]

[tex]w=0.78\frac{rad}{s}[/tex]

Calculation of the child's centripetal acceleration

We replaced [tex]w=0.78\frac{rad}{s}[/tex]and [tex]r=4.17m[/tex]in the equation (1)

[tex]a_{c} =0.78^{2} *4.17[/tex]

[tex]a_{c} =0.6084*4.17[/tex]

[tex]a_{c} =2.53\frac{rad}{s^{2} }[/tex]

Answer: The magnitude of the child's centripetal acceleration is [tex]2.53\frac{m}{s^{2} }[/tex]