An athlete whirls a 6.33 kg hammer tied to the end of a 1.4 m chain in a simple horizontal circle where you should ignore any vertical deviations. The hammer moves at the rate of 0.732 rev/s. What is the centripetal acceleration of the hammer? Assume his arm length is included in the length given for the chain. Answer in units of m/s 2 .

Respuesta :

Answer:

29.6 m/s^2

Explanation:

The centripetal acceleration is given by:

[tex]a=\omega^2 r[/tex]

where

[tex]\omega[/tex] is the angular velocity

r is the radius of the circular path

In this problem, we have the following data:

radius: r = 1.4 m

While we have to convert the angular velocity from rev/s to rad/s:

[tex]\omega=0.732 \frac{rev}{s} \cdot 2 \pi \frac{rad}{rev}=4.6 rad/s[/tex]

Therefore, the centripetal acceleration is

[tex]a=(4.6 rad/s)^2(1.4 m)=29.6 m/s^2[/tex]

ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE