Question

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 .

Answer 1

29.6 m/s^2

Step-by-step explanation:

The centripetal acceleration is given by:

`a=\omega^2 r`

where

`\omega` 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:

`\omega=0.732 (rev)/(s) \cdot 2 \pi (rad)/(rev)=4.6 rad/s`

Therefore, the centripetal acceleration is

`a=(4.6 rad/s)^2(1.4 m)=29.6 m/s^2`