A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in 1.0 s. a) What is the centripetal acceleration of the ball? b) What is the tension in the string?

Respuesta :

Answers:

a) [tex]0.5 m/s^{2}[/tex]

b) [tex]1.5 N[/tex]

Explanation:

a) The centripetal acceleration [tex]a_{c}[/tex] of an object moving in a uniform circular motion is given by the following equation:  

[tex]a_{c}=\omega^{2} r[/tex]  

Where:

[tex]\omega=1 \frac{rev}{s}[/tex] is the angular velocity of the ball

[tex]r=0.5 m[/tex] is the radius of the circular motion, which is equal to the length of the string

Then:

[tex]a_{c}=(1 \frac{rev}{s})^{2} 0.5 m[/tex]  

[tex]a_{c}=0.5 m/s^{2}[/tex] This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force [tex]F[/tex]) that is directed towards the center and is equal to the tension ([tex]T[/tex]) in the string:

[tex]F=T=m. a_{c}[/tex]

Where [tex]m=3 kg[/tex] is the mass of the ball

Hence:

[tex]T=(3 kg)(0.5 m/s^{2})[/tex]

[tex]T=1.5 N[/tex] This is the tension in the string